Fun with Maths and Physics by an.old.cathedralite

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In 1913 in Russian bookshops appeared a book by the outstanding educational-
ist Yakov Isidorovich Perelman entitled Physics for Entertainment. It struck the
fancy of the young who found in it the answers to many of the questions that in-
terested them.
Physics for Entertainment not only had an interesting layout, it was also im-
mensely instructive.
In the preface to the 11th Russian edition Perelman wrote: "The main objective
of Physics for Entertainment is to arouse the activity of scientific imagination,
to teach the reader to think in the spirit of the science of physics and to create in
his mind a wide variety of associations of physical knowledge with the widely
differing facts of life, with all that he normally comes in contact with."
Physics for Entertainment was a best seller.
 Ya. I. Perelman was born in 1882 in the town of Byelostok (now in Poland). In
 1909 he obtained a diploma of forester from the St. Petersburg Forestry Insti-
 tute. After the success of Physics for Entertainment Perelman set out to produce
 other books, in which he showed himself to be an imaginative popularizer of
 science. Especially popular were Arithmetic for Entertainment, Mechanics for
Entertainment,    Geometry for Entertainment,       Astronomy for        Entertainment,
Lively Mathematics, Physics Everywhere, and Tricks and Amusements.              Today
these books are known to every educated person in the Soviet Union.
 He has also written several books on interplanetary travel {Interplanetary Jour-
neys, On a Rocket to Stars, World Expanses, etc.).
The great scientist K. E. Tsiolkovsky thought highly of the talent and creative
genius of Perelman. He wrote of him in the preface to Interplanetary          Journeys:
 "The author has long been known by his popular, witty and quite scientific
works on physics, astronomy and mathematics, which are moreover written in a
marvelous language and are very readable."
Perelman has also authored a number of textbooks and articles in Soviet popular
science magazines.
In addition to his educational, scientific and literary activities, he has also devot-
ed much time to editing. So he was the editor of the magazines Nature and Peo-
ple and In the Workshop of Nature.
Perelman died on March 16, 1942, in Leningrad.
Many generations of readers have enjoyed Perelman's fascinating books, and
they will undoubtedly be of interest for generations to come.
Fun with Maths
 and Physics
    Brain Teasers

H. M. r i e p e j i b M a H

3AHMMATEJlbHbIE 3A/I.AMH M O n b l T b l

M 3/iarejihc I BO
«X(eTCKaH jiHTeparypa», MocKBa

Compiled by I. I. Prusakov
Translated from Russian by Alexander Repyev

Cover: I. Kravtsov, V. Stulikov
Artistic Book Design: I. Kabakov, V. Keidan, I. Kravtsov, D. Lion, S. Mukhin,   Yu. Perevezentsev,
L. Saksonov, A. Sokolov, V. Stulikov, R. Varshamov, Yu. Vashchenko

 First published 1984
 Second printing 1988


Printed in the Union of Soviet Socialist Republics

ISBN       5-03-000025-9                             © English translation, Mir Publishers, 1984
By the Way
6-7   For Young Physicists
A Sheet of Newspaper
8-9   Seventy-Five More Questions
      and Experiments               100
      on Physics
Optical Illusions
        Brain-Twisting Arrangements
10-11   and Permutations              164
Skilful Cutting and Connecting
22-13   n   Problems with Squares   1 8 4
Problems on Manual Work
14-15   Problems on Purchases
        and Prices              193
Weight and Weighing
16-17   Problems on Clocks and Watches
Problems on Transport
        Surprising Calculations
18-19                             215
20-21   Problems from Gullivers Travels
Stories about Giant Numbers
22-23   Tricks with Numbers
Merry Arithmetic
24-25        Count
         Fast Reckoning

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26-27        Magic Squares
     Arithmetic Games and Tricks
28-29         With a Stroke of the Pen

     Geometric Recreations
30-31   Without a Tape-Measure
           Simple. Tricks    I C O
           and Diversions
           Drawing Puzzles
27    O
32-33      By the Way

           Scissors and Paper
           Three pieces from one cutm Placing a strip on an edge• Charmed
           rings • Unexpected results of cutting • Paper chain • Thread yourself
           through a sheet of paper.
           Perhaps you think, as I once did, that there are some
           unnecessary things in this world. You're quite
           mistaken: there is no junk that might not be of help
           sometime for some purpose. What is useless for one
           purpose, comes in nandy for another, and what is
           useless for business might be suitable for leisure.
             In the corner of a room being repaired I once came
           across several used postcards and a heap of narrow
           paper strips that had been trimmed from wall paper.
           "Rubbish for the fire," I thought. But it turned out that
           even with this junk one can interestingly amuse oneself.
           My elder brother Alex showed me some things you
           could do with them.
             He started with the paper strips. Giving me a piece
           of strip about 30 centimetres long, he said: "Take
           a pair of scissors and cut the strip in three..."
             I was about to cut but Alex stopped me: "Wait a bit,
           I haven't finished yet. Cut it into three with one cut of
           the scissors."
             This was more difficult. I tried one way and then
           another, and then began to think my brother had posed
           a virtually impossible problem. Eventually it occurred
           to me that it was absolutely intractable.

Figure 1

             "You're pulling my leg," I said, "It's impossible."
             "Well, think again, maybe you'll work it out."
             "I have worked it out that the problem has no
             "Too bad. Let me."
             Brother took the strip, folded it in two and cut it in
           the middle to produce three pieces.
             "You see?"
m          38-39 By the Way
-&            "Yes, but you've folded the strip."
              "Why didn't you?"
              "You didn't say I could."
              "But I didn't say you couldn't either. Simply agree
           that you didn't see the solution."
              "All right. Give me another problem. You won't
           catch me out again."
Figure 2      "Here is another strip, put it on its edge."
              "So that it stands or falls?" I asked suspecting a trap.
              "Standing, of course. If it falls, it will mean that it
           was laid, not stood on edge."
              "So that it stands... on its edge," I muzed, and it
           suddenly occurred to me that I could bend the strip.
           I did so and put it on the table.
              "There, standing on its edge! You didn't say
           I couldn't bend it!" I said triumphantly.
              "More of your problems, please."
              "As you like. You see I've glued the ends of several
           strips and produced paper rings. Take a red-and-blue
           pencil and draw a blue line all along the outside of this
           ring and a red along the inside."
              "And then?"
              "That's all."
              A silly job, but somehow it didn't quite come off.
           When I had joined up the ends of the blue line and
           wanted to do the red, I found to my surprise I had
           absent-mindedly drawn the blue line on both sides of
           the ring.
              "Give me another," I was embarrassed, "I've
           accidentally spoiled the first one."
              But the second was a failure, too. I even didn't notice
           how I had drawn the both sides.
              "Some delusion! Again. Give me another."
              "You are welcome."
              Well, what do you think? Again, both sides appeared
           blue! There was no room for the red.
              I was upset.
              "Such a simple thing and you can't do it," smiled
           brother. "Just look."
              He took a paper ring and swiftly drew a red line all
           round the outside and a blue one on the inside.
              Having received a fresh ring, I started as carefully as
           possible to draw the line along one side, trying very
           hard not to go over to the other side somehow, and...
           joined up the line. Dear me! Both sides again. About to
           weep, I in bewilderment glanced at my brother, and
34-35      By the Way

           only then I guessed from his grin that something was
              "Well, you just... Is it a trick?" I asked.
              "The rings are magic."
              "What magic? Just rings. You've just fixed up
              "Try to make something else with these rings. For
           example, can you cut this ring to get two thinner
              "Nothing special."
              Having cut the ring, I was about to demonstrate two
           thin rings I had got when I noticed, much to my
           surprise, that I had in my hands only one long ring, not
           two smaller ones.
              "Okay, where are your two rings?" Alex asked
              "Another ring, I'll try again."
              "Why? Just cut the one you've got."
              I did. This time I had two rings, no kidding. But
           when I wanted to separate them, it turned out that it
           was impossible to disentangle them for they were linked
           together. Brother was right, the ring was enchanted all
              "The trick is very simple," my brother explained,
           "You can make such unusual rings for yourself. The
           key thing is that before you glue the ends of the paper
           strip twist one of them like this (Figure 3)."
              "Is it all because of that?"
              "Yes! Sure, I used an ordinary ring... It'll be even
Figure 3   more interesting, if the end is twisted twice, not just
              Before my very eyes Alex prepared a ring in this way
           and handed it to me.
              "Cutting along the middle," he said, "and see what
              I did and got two rings but one now went through
           the other. So funny, it was impossible to take them
              I prepared three more rings for myself and obtained
           three more pairs of inseparable rings.
              "What would you do," my brother asked again, "if
           you had to connect all four pairs of rings to form one
           long open-ended chain?"
               "Oh, this is simple: cut one ring in each pair, and
           glue them together again."
               Alex enquired, "So, you would cut three of the
           38-39 By the Way
             "Of course."
             "But what if you cut less than three?"
             "We have four pairs of rings, how can you possibly
           connect them by only breaking two rings? Impossible!"

Figure 4

              I was dead sure.
              In answer, my brother took the scissors, cut both
           rings in one pair and with them connected the
           remaining three pairs. Lo and behold! a chain of eight
           rings. Ridiculously simple! N o trick in this and I could
           only be surprised why such a simple idea hadn't
           occurred to me.
              "Enough of these paper rings. You've got some old
           postcards over there, it seems. Let's have some fun with
           them, too. For instance, try and cut in a card the
           largest hole you can."
              I punched the card with my scissors and carefully
           cut a rectangular hole in it, leaving only a narrow edge.
              "This is a hole among holes! A larger one is
           impossible!" I contentedly showed the result of my job
           to Alex.
              Of course, he had another opinion.
              "The hole is too small. Just enough for a hand to go
              "You'd like it to be large enough for a head?"
           I retorted acidly.
              "The head and the body. So that you could thread
           all yourself through it. That would be some hole."
              "Ha-ha! Do you really want to get a hole larger than
           the paper itself?"
              "Exactly. Many times larger."
              "No trick will help you here. What is impossible is
              "And what is possible is possible," said Alex and set
           out to cut.
              Confident that he was joking, nevertheless, I watched
           him curiously. He bent the card in two, then drew two
           lines with a pencil near the long edges of the bent
           postcard and made two incisions near the other two
              Next he cut the bent edge from point A to point
38-39       By the Way

            B and began to make a lot of cuts next to each other as
            shown in Figure 5.
              "Finished," proclaimed my brother.
              "Why? I see no hole!"
Figure 5

              "Take another look."
              And Alex expanded the paper. Just imagine: it
            developed into a long-long chain that he easily threw
            over my head. It fell to my feet, zigzagging about me.
              "How can you get through such a hole? What do
            you say to that?"
              "Big enough for two!" I said with admiration.
              At that Alex finished his tricks, promising to treat
            me next time to a whole heap of new ones, this time
            only with coins.

            Tricks with Coins
           A visible and invisible coin • A bottomless glass • Where has the coin
           goneArranging        coins0 Which hand holds the two-pence     piece?•
           Shi/ting coins • An Indian legend • Problem solutions.
            "Yesterday you promised to show me a trick with
            coins," I reminded my brother at breakfast.
              "Tricks? First thing in the morning? Hm, all right.
            Then empty the washing bowl."
              Alex put a silver coin on the bottom of the empty
              "Look into the bowl without moving from your place
            and without leaning over. See the coin?"
              Alex pushed the bowl a bit farther away from me
            until I couldn't see the coin any more since it was
            shielded by the side of the bowl.
              "Sit still, don't move. I pour water into the bowl.
            What has happened to the coin?"
              "It's visible again, as if it's been lifted up together
            with the bottom. Why?"
              My brother sketched the bowl with the coin in it on
            a sheet of paper, and then everything became clear to
           38-39 By the Way

           me. While the coin was at the bottom of the dry bowl,
           no ray of light could come from the coin because light
Figure 6   travels in straight lines and the opaque sides of the
           bowl were just in the way. When the water was added,
           the situation changed since light rays coming from
           water into air get bent (physicists say "refracted") and
           now can slid over the bowl edge and come into my
           eyes. However, we are used to seeing things only at
           a place where straight rays come from and this is why
           I mentally placed the coin somewhat higher than where
           it really was, that is along a continuation of the
           refracted ray. So it seemed to me that the bottom had
           risen with the coin.
              "I advise you to remember this experiment," my
           brother added. "It will be useful when bathing. In
           a shallow place where you can see the bottom never
           forget that you see it higher than really is. And
           substantially so, for water appears to be shallower by
           about a quarter of its real depth. Where the actual
           depth is 1 metre, say, the apparent depth is only 75
           centimetres. Bathing children often get into a trouble
           for this reason: relying on the deceptive appearance,
           they usually underestimate depth."
              "I noticed that when you float slowly in a boat over
           a place where the bottom is visible, it appears that the
           greatest depth is just under the boat and it's much
           shallower everywhere else. But shift to another place
           and again everywhere is shallow and beneath the boat
           is deep. It seems as if the deepest place travels with the
           boat. Why?"
              "Now you can understand that easily. The point is
           that the rays coming straight up out of the water
           change direction least of all, thus the bottom there
           appears to be less elevated than the places which send
           oblique rays to our eyes. Naturally, the deepest place
           appears to us to lie just beneath the boat even if the
           bottom were perfectly flat. But now let's do quite
           another experiment."
              Alex filled a glass with water right up to the brim.
              "What do you think will happen if now I drop
           a two-pence piece into the glass?"
             "The water will overflow, of course."
             "Let's try."
             Carefully, without jerking, my brother lowered the
           coin into the brimful glass. N o t a drop overflowed.
             "Now let's put in another coin," he said.
             I warned him, "Now it's sure to overflow."
        By the   Way

           And I was mistaken: in the full glass there was room
        for the second coin, too. A third and a fourth coin
        followed each other into the glass.
           "What a bottomless glass!" I exclaimed.
           Alex silently and cooly kept on lowering one coin
        after another into the glass. A fifth, sixth, seventh time
        coins fell onto the b o t t o m - n o overflowing. I couldn't
        believe my eyes and was impatient to find out the
           But my brother took his time to explain, he was still
        carefully dropping coins and only stopped at the 15 th
        two-pence piece.
           "Well, that'll do," he said at last, "Take a look, the
        water has bulged up at the glass's edge."
           Indeed, the water had bulged above the edge by
        about the thickness of a match, sloping down at the
        edges as if it were in a transparent bag.
           Alex went on to say, "The answer lies in the bulging.
        This is where the water is that was expelled by the
           "Fifteen coins have displaced so little water?" I was
        astounded. The stack of 15 two-pence pieces is rather
        high but here is only a thin layer, just thicker than
        a penny."
           "Take its area into consideration, not only the
        thickness. The layer may be not thicker than
        a two-pence piece, but how many times larger is it
           I gave some thought to i t - t h e glass was about four
        times wider than a two-pence.
           "Four times wider and the same thickness." I went
        on to conclude, "The layer is only four times larger
        than a two-pence. The glass could only receive four
        coins, but you've already put in 15 and plan, it seems,
        to add some more. Where's the room?"
           "Your calculation is wrong. If a ring is four times
        larger across than another, its surface area will be 16
        times larger, not four."
           "Well, I never!"
           "You should have known it. How many square
        centimetres are there in a square metre? One
           "No, 100 times 100 which is 10,000."
           "You see. With rings the same rule holds: if a ring is
        two times wider than another, it has four times the
        surface area, three times wider-nine times the area,
        four times wider-16 times the area, and so on. So, the
38-39 By the Way
volume of the buldge above the brim is 16 times larger
than that of a two-pence piece. You see now where all
the room is in the glass. It has even more room because
the water can rise up about two two-pence pieces
   "Could you really put 20 coins into the glass?"
   "Even more, if only you dip them carefully, without
   "I wouldn't ever have believed that a brimful glass
could have enough room for so many coins."
   I had to believe it though when I saw the heap of
coins inside the glass with my own eyes.
   "Now, can you place 11 coins into 10 saucers so that
there is only one coin in each saucer?" the brother
   "Saucers with water?"
   "With or without water, as you please," he laughed,
setting 10 saucers in a row.
   "Another physics experiment?"
   "No, psychological. On with the job."
   "Eleven coins in 10 saucers, and one in each... No,
I can't," I gave up at once.
   "Go ahead, I'll help you. We'll place the first coin in
the first saucer and the 11th as well, just for a time."
   I did as he said, waiting in bewilderment. What is
going to follow?
   "Two coins? Well, the third coin goes into the
second saucer. The fourth into the third saucer, the fifth
into the fourth, and so forth."
   When I had placed the 10th coin into the ninth
saucer I was surprised to see that the 10th saucer was
   Alex said, "Now that's where we'll place the 11th
coin that we put tentatively into the first saucer." He
took the extra coin from the first saucer and placed it
into the 10th saucer.
   Now 11 coins were lying in 10 saucers, one in each...
   Brother swiftly collected the coins not caring to
explain the trick to me.
   "Just think. That'll be both more interesting and
more useful than getting ready-made solutions."
   And ignoring my pleads he gave me a fresh problem.
   "Here are six coins. Arrange them in three rows so
that there are three coins in each."
   "That takes nine coins."
   "Everyone can do it with nine. No, do it just with
40—41      By the Way

Figure 7

              "Now again that's something impossible."
              "You're too quick to give up. Look, it's simple."
              "There are three rows here, with three coins in each,"
           he explained.
              "But the rows criss-cross."
              "Perhaps, but did I say that they mustn't?"
              "If I'd known that this was allowed, I'd have guessed
           for myself."
              "Well, then, guess how to solve the problem in
           another way. But not now, sleep on it. Here are three
           more problems in the same vein. The first one: arrange
           nine coins in ten rows with three coins in each. The
Figure 8
           second: arrange ten coins in five rows with four coins
           in each. The third problem is as follows. I draw
           a square divided into 36 smaller squares. Now try to
           arrange 18 coins with one in each small square so that
           in each row and column there are three coins... Aha,
           I've just remembered one more trick with coins. Take
           into one hand a 5 pence, into the other a 10 pence, but
           don't tell me which coin is in which hand. I'll figure it
           out. Only do the following mental arithmetic: double
           what's in the right hand and treble what's in the left,
           and then add the results. Ready?"
              "What's the final result, odd or even?"
              "The 10 is in the right and the 5 in the left hand,"
           Alex proclaimed at once and was right on target.
              We repeated it once more. The result was even and
           my brother said without mistake that the 10 was in my
           left hand.
              "About this problem also think at leisure," he said,
           "and finally, I'll show you a fascinating game with
           counters. I've just made some counters by cutting out
           differently sized disks from a sheet of cardboard. The
           biggest counter is 5 centimetres in diameter, the next
           biggest 4 centimetres, and so on down to the smallest
           which is 1 centimetre in diameter."
           38-39 By the Way
              He put three saucers side by side, and put a stack of
           counters onto the first saucer: so that the 5 counter
           went on the bottom on top of that was the 4 counter,
           and so on down to the 1 counter on top of the stack.
             "The whole stack of five counters is to be transferred
           onto the third saucer but you have to observe the
           following rules. Rule number 1: each time move
           1 counter only. Rule number 2: never put a larger

Figure 9

             C5D CcS^CcT)
            counter onto a smaller one. Rule number 3: counters
            may be placed temporarily onto the middle saucer but
            still observing the first two rules and the counters must
            end up on the third saucer in the initial order. The
           rules are simple as you can see. Now, go ahead."
               I started. First I placed the 1 counter onto the third
           saucer, the 2 counter onto the middle one... and
           stopped. Where should the 3 counter go? It was larger
           than both the 1 and 2 counters.
               "Well, then," my brother prompted, "Place the
            1 onto the middle saucer, then the third saucer will be
           vacant for the 3.
               I did so. Now a further predicament. Where was I to
           place the 4? Accidentally, I hit upon an idea: first
           1 transferred the 1 onto the first saucer, the 2 onto the
           third, and next put the 1 onto the third as well. Finally,
           after a long series of move I succeeded in transferring
           the 5 from the first saucer and ended up with the whole
           stack on the third saucer.
               "How many transfer did you make in all?" asked my
           brother okaying my job.
               "Didn't count."
               "Well let's count then. After all, it's interesting to find
           the least number of moves that could lead to the goal.
           If our stack included only two counters, not five, the
           2 and the 1, how many moves would be required?"
               "Three: the 1 onto the middle saucer, the 2 onto the
           third one and then the 1 onto the third."
               "Right. Add one more counter, the 3, and count how
           many moves you need to transfer the stack. We'll
           proceed as follows: first we transfer the two smaller
           coins onto the middle saucer one after the other. This
           takes, as we already know, three moves. We then
           transfer the 3 onto the vacant third saucer-one more
42^3   38-39 By the Way

       move. Next we transfer both counters from the middle
       saucer, too, onto the third o n e - t h r e e more moves. The
       total is 3 + 1 + 3 = 7."
          "For the four counters, let me count for myself. At
       first I transfer the three smaller counters onto the
       middle saucer-seven moves; then the 4 goes onto the
       third s a u c e r - o n e move, and now the three smaller
       coins go onto the third saucer-seven more moves.
       Thus I get 7 + 1 + 7 = 15."
          "Splendid. And for the five counters?"
          "15 + 1 + 15 = 31."
          "Well, you got it right. But I'll show you a way to
       simplify the procedure. Note that the numbers
       involved-3, 7, 15, 3 1 - a l l represent the product of
       several twos minus one. Look!" And Alex wrote out
       the following table.
        3=    2x2-1
        7=    2x2x2-1
       15 =    2x2x2x2-1
       31 =    2x2x2x2x2-1
         "I see, the number of the counters to be transferred
       equals the number of twos in the product. Now, I could
       calculate the number of moves for any stack of
       counters. For instance, for seven counters: it's
       2 x 2 x 2 x 2 x 2 x 2 x 2 - 1 = 1 2 8 - 1 = 127.
          "You've thus mastered this ancient game. You need
       only know one practical rule which is if the stack
       contains an odd number of counters the first counter is
       transferred onto the third saucer, if its's even, it goes
       onto the middle saucer."
          "You say it's an ancient game, didn't you invent it
          "No, I only applied it to counters. But the game as
       such has a very ancient origin and apparently came
       from India where there is a marvellous legend
       associated with it. It says that in the town of Benares is
       a sanctuary into which the Indian god Brahma, as he
       was creating the world, installed three diamond sticks
       and put on one of them 64 golden rings with the largest
       at the bottom and each of the rest being smaller than
       the one beneath it. The priests of the sanctuary were
       obliged ceaselessly to transfer these rings from one stick
       to another using the third as an auxiliary and observing
       the rules of our game that is to move one ring at a time
            38-39 By the Way
            and not to place it onto a smaller one. The legend has
            it that when all the 64 rings have been transferred the
            end of the world will come."
               "Oh, it means the world should've perished long
               "Perhaps, you think transferring 64 rings won't take
            much time?"
               "Of course. Allowing a second per move, you can
            make 3600 transferrings in an hour."
               "And about 100,000 in 24 hours. In 10 days,
            a million moves. A million would be enough, I think, to
            transfer even a thousand, not 64 rings."
               "You are mistaken. To handle the 64 rings would
            take as much as 500,000,000,000 years!"
               "But, why? After all, the number of moves is only
            equal to the product of 64 twos, which amounts to..."
               "'Only' upwards of 18,000,000,000,000,000,000."
               "Wait a bit, I'll now multiply and check."
               "Splendid. While you do your multiplying, I'll have
            time to go to tend to my business," said brother and
               I first found the product of 16 twos, then multiplied
            the result by itself. A tedious job, but I was patient and
            worked it out to the end. I obtained the number
               Thus my elder brother was right...
               I mustered up courage and set about the problems he
            had set to me to solve on my own. They didn't turn out
            to be all that difficult, some were even rather easy. The
            business of 11 coins in 10 saucers appeared ridiculously
            simple: we put the first and eleventh coins into the first
            saucer, next we put the third coin into the second
            saucer, the fourth coin into the third saucer, and so
            forth. But what about the second coin? It was ignored
            and that was the trick. The idea behind guessing which
            hand had the 10 pence coin was also simple. Doubling
            5 gives an even number but trebling it gives an odd one
Figure 10
44-45       By the Way

Figure 11
                    ©©                                 ©
                    ©                                  ©©
                       ©© ©
                    ©   © ©
                      ©© ©
            whereas multiplying 10 always gives an even number.
            Therefore, if the total was even, then the 5 had been
            doubled, i. e. it must have been in the right hand, and if
            the total was odd, it is clear that the 5 must have been
            trebled, i.e. been in the left hand. The solutions to the
            problems on the coin arrangements are clear from the
            accompanying drawings (Fig. 10).
               Finally, the problem with coins in the small squares
            works out as shown in Fig. 11. The 18 coins are
            arranged in the square with 36 small squares and giving
            three coins in each row.

            Wandering in a Maze
            Wandering in a maze • People and rats 0 Right- and left-hand rule •
            Mazes in ancient times • Tournefort in a cave • Solution of the maze

            "What are you laughing at in your book? A funny
            story?" Alex asked me.
              "Yes, it's Three Men in a Boat by Jerome."
               "I remember it had me in stitches! Where are you?"
               "Where the crowd of people is wandering about in
            a garden maze, looking for a way out."
               "An interesting story. Read it again for me, please."
               So I read the story aloud from the very beginning:
               "Harris asked me if I'd ever been in the maze at
            Hampton Court. He said he went in once to show
38-39 By the Way
somebody else the way. He had studied it up in a map,
and it was so simple that it seemed foolish-hardly
worth the twopence charged for admission. Harris said
he thought that map must have been got up as
a practical joke because it wasn't a bit like the real
thing, and only misleading. It was a country cousin that
Harris took in. He said: 'We'll just go in here, so that
you can say you've been, but it's very simple. It's
absurd to call it a maze. You keep on taking the first
turning to the right. We'll just walk round for ten
minutes, and then go and get some lunch.'
    "They met some people soon after they had got
inside, who said they had been there for three-quarters
of an hour, and had had about enough of it. Harris
told them they could follow him, if they liked; he was
just going in, and then should turn round and come
out again. They said it was very kind of him, and fell
behind, and followed.
    "They picked up various other people who wanted to
get it over, as they went along, until they had absorbed
all the persons in the maze. People who had given up
all hopes of ever getting either in or out, or of ever
seeing their home and friends again, plucked up cour-
age, at the sight of Harris and his party, and joined the
procession, blessing him. Harris said he should judge
there must have been twenty people following him, in
all; and one woman with a baby, who had been there
all the morning, insisted on taking his arm, for fear of
losing him.
    "Harris kept on turning to the right, but it seemed
a long way, and his cousin said he supposed it was
a very big maze.
    " ' O h , one of the largest in Europe,' said Harris.
    "'Yes, it must be, replied the cousin, because we've
 walked a good two miles already.'
    "Harris began to think it rather strange himself, but
he held on until, at last, they passed the half of a penny
bun on the ground that Harris's cousin swore he had
 noticed there seven minutes ago. Harris said, 'Oh,
 impossible!' But the woman with the baby said, 'Not at
 all,' as she herself had taken it from the child, and
 thrown it down there, just before she met Harris. She
 also added that she wished she never had met Harris,
 and expressed an opinion that he was an impostor.
 That made Harris mad, and he produced his map, and
 explained his theory.
    "'The map may be all right enough, said one of the
38-39   By the   Way

        party, if you know whereabouts in it we are now.'
           "Harris didn't know, and suggested that the best
        thing to do would be to go back to the entrance, and
        begin again. For the beginning again part of it there
        was not much enthusiasm; but with regard to the
        advisability of going back to the entrance there was
        complete unanimity, and so they turned, and trailed
        after Harris again, in the opposite direction. About ten
        minutes more passed, and then they found themselves
        in the centre.
           "Harris thought at first of pretending that that was
        what he had been aiming at; but the crowd looked
        dangerous, and he decided to treat it as an accident.
           "Anyhow, they has got something to start from then.
        They did know where they were, and the map was once
        more consulted, and the thing seemed simpler than
        ever, and off- they started for the third time.
           "And three minutes later they were back in the centre
           "After that they simply couldn't get anywhere else.
        Whatever way they turned brought them back to the
        middle. It became so regular at length, that some of the
        people stopped there, and waited for the others to take
        a walk round, and come back to them. Harris drew out
        his map again, after a while, but the sight of it only
        infuriated the mob, and they told him to go and curl
        his hair with it. Harris said that he couldn't help feeling
        that, to a certain extent, he had become unpopular.
           "They all got crazy at last, and sang out for the
        keeper, and the man came and climbed up the ladder
        outside, and shouted out directions to them. But all
        their heads were by this time, in such a confused whirl
        that they were incapable of grasping anything, and so
        the man told them to stop where they were, and he
        would come to them. They huddled together, and
        waited; and he climbed down, and came in.
           "He was a young keeper, as luck would have it, and
        new to the business; and when he got in, he couldn't
        get to them, and then he got lost. They caught sight of
        him, every now and then, rushing about the other side
        of the hedge, and he would see them, and rush to get to
        them, and they would wait there for about five minutes,
        and then he would reappear again in exactly the same
        spot, and ask them where they had been.
           "They had to wait until one of the old keepers came
        back from his dinner before they got out."
           "They were a bit dense," I said, "To have a plan and
           Bay the Way

           not to find the way out."
             "Do you think you'd find at once?"
             "With a plan? Certainly!"
             "Just wait. It seems to me I've got the plan of that
           maze," Alex said and began to delve in his bookcase.
             "Does this maze really exist?"
             "Hampton Court? Of course, it's near London. Been
           in existence for two centuries... Found at last. Just as
Figure K

           I said Plan of the Maze at Hampton Court. It seems
           rather small, this maze, only 1,000 square metres."
              My brother opened the book at a page showing
           a small plan.
              "Imagine you're here in the central area of the maze
           and want to get out, which way would you go to get to
           the exit? Sharpen a match and use it to show the way."
              I pointed the match at the centre of the maze and
           bravely drew it along the winding paths of the maze,
           but the whole affair appeared to be more involved than
           I had expected. Having wandered a little round about
           the plan I came... back to the central area, just as
           Jerome's characters had, the ones I'd just made fun of!
              "You see, the plan is no use. But rats solve the task
           without any plan."
              "Rats? What rats?"
              "The ones described in this book. Do you think this
           is a treatise on garden design? No, this book is about
           the mental abilities of animals. To test the intelligence
           of animals, scientists make a small plaster model of
           a maze and put the animals to be tested into it. The
           book says that rats can find their way about a plaster
           maze of Hampton Court in only half an hour and that
           is faster than the people in Jerome's book."
   38-39    By the Way

               "Judging from the plan, the maze doesn't seem to be
            very difficult. You would never think that it's so
               "There's a simple rule. If you know it, you can safely
            enter any maze without any fear of getting lost."
               "Which rule?"
               "You should follow the paths touching its wall with
            your right hand, or left for that m a t t e r - i t makes no
            difference. But with one hand, all the time."
               "Just this?"
               "Yes. Now try and use the rule in reality, mentally
            wandering about the plan."
               I ran my match along the paths, being guided by the
            rule. Truly, I soon came from the entrance to the centre
            and back again, to the exit.
               "A beautiful rule."
               "Not really," Alex objected, "The rule is good so long
            as you simply don't want to be lost in a maze, but it's
            no good if you want to walk along all of its paths
            without exception."
               "But I've just been in all the alleys on the plan.
             I didn't miss one."
               "You are mistaken. Had you marked with a dash line
             the way you went, you'd have found that one alley
             wasn't covered."
                "Which one?"
Figure 13

              "I've marked it with a star on this plan (Fig. 13).
            You haven't been down this alley. In other mazes the
            rule would guide you past large sections of it so that
            even though you'd find your way out safely, you
            wouldn't see much of it."
              "Are there many different kinds of maze?"
              "A lot. Nowadays they are only in garden and parks
            and you wander around in the open air between high
            green walls of hedge, but in ancient times they used to
            put mazes inside large houses or dungeons. That was
4 —97J
38-39       By the Way

            done with the cruel aim of dooming the unhappy
            people thrown into them to wander hopelessly about
            the intricate tangle of corridors, passages and halls,
            eventually leading them to starve to death. One such,
            for instance, was the proverbial maze on the island of
            Crete and the legend has it that an ancient king called
            Minos had it built. Its paths were so tangled that its
            own creator, a man called Daedalus, allegedly couldn't
            find his way out of it," brother continued, "The aim of
            other ancient mazes was to guard the tombs of kings,
            to protect them from robbers. A tomb was located at
            the centre of a maze so that if a greedy seeker after
            buried treasure even succeeded in reaching it, he
            wouldn't be able to find his way o u t - t h e grave of the
            king would become his grave, too."
               "Why didn't they use the rule for walking round
Figure 14   mazes you've just told me about?"
               "For one thing, apparently, in ancient times nobody
            knew about the rule. For another, I've already told you
            that it doesn't always let you visit every part of the
            maze. A maze can be contrived in such a way that the
            user of the rule will miss the place where the treasure is
               "But is it possible to make a maze from which there
            is no escaping? Of course, someone who enters it using
            your rule will get out eventually, but suppose a man is
            put inside and left there to wander?"
               "The ancients thought that when the paths of a maze
            are sufficiently tangled, it would be absolutely
            impossible to get out of it. This isn't true because it can
            be proved mathematically that inescapable mazes
            cannot be built. Not only that but every maze has an
            escape and it is possible to visit every corner without
            missing one and still escape to safety. You only need to
            follow a strict system and take certain precautions. Two
            centuries ago the French botanist Tournefort dared to
            visit a cave in Crete which was said to be an inescapable
            maze because of its innumerable paths. There are
            several such caves in Crete and it may be possible that
            they gave rise to the ancient legend about the maze of
            King Minos. What did the French botanist do in order
            not to be lost? This is what his fellow-countryman, the
            mathematician Lucas, said about it."
               My brother took down from the bookcase an old
            book entitled Mathematical Amusements and read aloud
            the following passage (I copied it later):
               "Having wandered for a time with our companions
50-51       By the Way

            about a network of underground corridors, we came to
            a long wide gallery that led us into a spacious hall deep
            in the maze. We had counted 1460 steps in half an hour
            along this gallery, deviating neither right nor left... On
            either side there were so many corridors that one
            would be bound to get lost there unless some necessary
            precautions were taken. But as we had a strong desire
            to be out of the maze, we saw to it to provide for our
               "First, we left one of our guides at the entrance to
            the cave, having instructed him to call for the people
            from a neighbouring village to rescue us should we not
            return by night fall. Second, each of us had a torch.
            Third, at every turn which, it seemed, might be difficult
Figure 15   to find later we attached numbered papers to the wall.
            Finally, one of our guides put on the left side bunches
            of blackthorn he had prepared beforehand, and the
            other side of the path he sprinkled with chopped straw,
            which he carried in a bag."
               Alex finished reading and said, "All these laborious
            precautions might not seem all that necessary to you.
            In the times of Tournefort, however, there was no other
            way since the problem of mazes had not yet been
            solved. These days the rules for walking around mazes
            have been worked out that are less burdensome but no
            less reliable than his precautions."
               "Do you know these rules?"
               "They aren't complicated. A first rule is that when
            you walk into a maze, follow any path till you reach
Figure 16   a dead end or a crossing. If it is a dead end, return and
            place two stones at the exit which will indicate that the
            corridor has been passed twice. At a crossing, go fur-
            ther down any corridor but mark each time you go
            down it with a stone the way you have just passed and
            the way you are going to follow. A second rule states
            that having arrived along a fresh corridor at a crossing
            that has earlier been visited (as seen by the stones), go
            back at once and place two stones at the end of the
            corridor. Finally, a third rule requires that having come
            to a visited crossing along a corridor that has already
            been walked, mark the way with a second stone and go
            along one of new corridors. If there doesn't happen to
            be such a corridor, take one whose entrance has only
            one stone (that is a corridor that has only been passed
            once). Abiding by these rules you can pass twice, that is
            there and back, every corridor of the maze without
            missing any corner and return back to safety. Here I've
38-39 By the Way
got several plans of mazes I've cut out at different times
from illustrated magazines (Figs. 14, 15, and 16). If you
wish, you can try and travel about them. I hope that
now that you know so much you shouldn't be in any
danger of getting lost in them. If you've enough
patience, you could actually make a maze like, say, the
Hampton Court one that Jerome mentioned - you
could construct it with your friends out of snow in the
52-53           %
                    For Young Physicists

            2       More Skilled Than    Columbus
                    A schoolboy once wrote in a composition: "Christo-
                    pher Columbus was a great man because he discovered
                    America and stood an egg upright." This young scholar
                    had thought both deeds equally amazing. On the
                    contrary, the American humorist Mark Twain saw
                    nothing special about Columbus discovering America:
                    "It would have been strange if he hadn't found it
                       The other feat the great navigator had performed is
                    not really all that marvellous. Do you know how
                    Columbus stood an egg upright? He simply pressed it
                    down onto a table crushing the bottom of the shell. He
                    had, of course, changed the shape of the egg. But how
                    can one possibly, stand an egg on end without changing
                    its shape, the navigator didn't know.
                       Meanwhile it is easier by far than discovering
                    America or even one tiny island. I'll show you three
                    methods: one for boiled eggs, one for raw eggs, and one
                    for both.
                       A boiled egg can be stood upright simply by spinning
                    it with your fingers or between your palms like a top.
                    The egg will remain upright as long as it spins. After
                    two or three trials the experiment should come out
                       This won't work if you try to stand a raw egg up-
                    right, you may have noticed that raw eggs spin poorly.
                    This, by the way, is used to distinguish a hard-boiled
                    egg from a raw one without breaking the shell. The
                    liquid contents of a raw egg is not carried along by the
                    spinning as fast as the shell and, therefore, sort of
                    damps the speed down. We have to look for another
                    way of standing eggs and one does exist. You have to
Figure 17           shake an egg intensely several times. This breaks down
                    the soft envelope containing the yolk with the result
                    that the yolk spreads out inside the egg. If you then
                    stand the egg on its blunt end and keep it this way for
                    a while, then the yolk, which is heavier than the white,
                    will pour down to the bottom of the egg and
                    concentrate there. This will bring the centre of mass of
                    the egg down making it more stable than before.
                       Finally, there is a third way of putting an egg up-
                    right. If an egg is placed, say, on the top of a corked
                    bottle and another cork with two forks stuck into it is
                    placed on the top as shown in Fig. 17, the whole
            For Young     Physicists

            system (as a physicist would put it) is fairly stable and
Figure 18   remains in equilibrium even if the bottle is slightly
            inclined. But why don't the egg and cork fall down?
            For the same reason that a pencil placed upright on
            a finger doesn't fall off when a bent penknife is stuck
            into it as shown. A scientist would explain: "The centre
            of mass of the system lies below the support." This
            means that the point at which the weight of the system
            is applied lies below the place at which it is supported.

            Centrifugal      Force
            Open an umbrella, put its end on the floor, spin it and
            drop a ball into it. The ball could be a balled piece of
            paper or handkerchief, or any other light and
            unbreakable thing. Something will happen you
Figure 19   probably wouldn't expect. The umbrella does not, as it
            were, desire to accept the present and the thing itself
            crawls up the edge and then flies off in a straight line.
               The force that threw the ball out in this experiment
            is generally called the "centrifugal force", although it
            would be more appropriate to dub it "inertia".
            Centrifugal force manifests itself when a body travels in
            a circle but this is nothing but an example of inertia
            which is the desire of a moving body to maintain its
            speed and direction.
               We come across centrifugal force more often than
            you might suspect. If you whirl a stone tied to a piece
            of string, you can feel the string become taut and seem
            to be about to break under the action of the centrifugal
Figure 20
            force. The ancient weapon for hurling stones, the sling,
            owes its existence to the force. Centrifugal force bursts
            a millstone, if it is spun too fast and is not sufficiently
            strong. If you are adroit enough, this force will help
            you to perform a trick with a glass from which the
            water doesn't escape, even though it is upside down. In
            order to do this you'll only have to swing the glass
            quickly above your head in a circle. Centrifugal force
            helps a circus bicyclist to do a "devil's loop". It is put
            to work. In the so-called centrifugal separators it
            churns cream; it extracts honey from honey-comb; it
            dries washing by extracting water in centrifugal driers,
            etc., etc.
               When a tram travels in a circular path, e.g. as it
            turns at a crossing, the passengers feel directly the
            centrifugal force that pushes them in the direction of
            the outer wall of the carriage. If the speed is sufficiently
54-55   For Young   Physicists

        large, the carriage could be overturned by the force if
        the outer rail wasn't laid a bit higher than the inner
        one: which is why a tram is slightly inclined inwards
        when it turns. It sounds rather unusual but an inclined
        tram is more stable than an upright one!
           But this is quite the case, though. A small experiment
        will help explain this to you. Bend a cardboard sheet to
        form a wide funnel, or better still take a conical bowl if
        available. The conical shield (glass or metallic) of an
        electrical lamp would be suitable for our purposes. Roll
        a coin (small metal disk, or ring) around the edge of
        any of these objects. It will travel in a circle bending in
        noticeably on its way. As the coin slows down, it will
        travel in ever decreasing circles approaching the centre
        of the funnel. But by slightly shaking the funnel the
        coin can easily be make roll faster and then it will move
        away from the centre describing increasingly larger
        circles. If you overdo it a bit, the coin will roll out.
           For cycling races in a velodrome special circular
        tracks are made and you can see that these tracks,
        especially where they turn abruptly have a noticeable
        slope into the centre. A cyclist rides along them in an
        inclined position (like the coin in the funnel) and not
        only does he not turn over but he acquires special
        stability. Circus cyclists used to amaze the public by
        racing along a steep deck. Now you can understand
        that there is nothing special about it. On the contrary,
        it would be a hard job for a cyclist to travel along
        a horizontal track. For the same reason a rider and his
        horse lean inwards on a sharp turn.
           Let's pass on from small to large-scale phenomena.
        The Earth, on which we live, rotates and so centrifugal
        force should manifest itself. But where and how? By
        making all the things on its surface lighter. The closer
        something is to the Equator, the larger the circle in
        which it moves and hence it rotates faster, thereby
        losing more of its weight. If a 1-kg mass were to be
        brought from one of the poles to the Equator and
        reweighed using a spring balance, the loss in weight
        would amount to 5 grammes. That, of course, is not
        very much of a difference, but the heavier a thing, the
        larger the difference. A locomotive that has come from
        Stockholm to Rome loses 60 kg, the weight of an adult.
        A battle ship of 20,000-tonne displacement that has
        come from the White Sea to the Black Sea will have
        lost as much as 80 tonnes, the weight of a locomotive!
           Why does it happen? Because as the globe rotates, it
54-55         For Young   Physicists

tries to throw everything off its surface just like the
umbrella in our earlier experiment. It would succeed
were it not for the terrestrial attraction that pulls
everything back to the Earth's surface. We call this
attraction "gravity". The rotation cannot throw things
off the Earth's surface, but it can make them lighter.
   The faster the rotation, the more noticeable the
reduction in weight. Scientists have calculated that if
the Earth rotated 17 times faster, things at the Equator
would lose their weight completely to become
weightless. And if it rotated yet quicker, making, say,
one turn every hour, then the weight lessness would
extend to the lands and seas farther away from the
   Just imagine things losing their weight. It would
mean there would be nothing you could not lift, you
would be able to lift locomotives, boulders, cannons
and warships as easily as you could a feather. And
should you drop t h e m - n o danger, they could hurt
nobody since they wouldn't fall down at all, but would
float about in mid-air just where you'd let go of them.
If, sitting in the cabin of an airship, you wanted to
throw something overboard, it wouldn't drop, but
would stay in the air. What a wonder world it would
be. So you could jump as high as you've never
dreamed, higher than sky-scrapers or the mountains.
But remember, it would be easy to jump up but diffi-
cult to return back to ground. Weightless, you'd never
come back on your own.
   There would also be other inconveniences in such
a world. You've probably realized yourself that
everything, whatever its size, would, if not fixed, rise up
due to the slightest motion of air and float about.
People, animals, cars, carts, ships-everything would
move about in the air disorderly, breaking, maiming
and destroying.
   That is what would occur if the Earth rotated sig-
nificantly faster.

Ten   Tops
The accompanying figures show 10 types of tops. These
will enable you to do a number of exciting and
instructive experiments. You don't need any special
skill to construct them so you can make them yourself
without any help or expense.
  This is how the tops are made:
56-57       54-55         For Young   Physicists

Figure 21      1. If a button with five holes comes your way, like
            the one shown in the figure, then you can easily make
            it into a top. Push a match with a sharpened end
            through the central hole, which is the only one needed,
            wedge it in and then... the top is ready. It will rotate
            on both the blunt and pointed end, you only need to
            spin it as usual by twisting the axle between your
            fingers and dropping the top swiftly on its blunt end. It
            will spin rocking eccentrically.
               2. You could do without a button, a cork is nearly
            always at hand. Cut a disk out of it, pierce the disk
            with a match, and you have top No. 2 (Fig. 22).
               3. Figure 23 depicts a rather unusual t o p - a walnut
Figure 22   that spins on the pointed end. To turn a suitable nut
            into a top just drive a match into the other end, the
            match being used for spinning the top.
               4. A better idea is to use a flat wide plug (or the
            plastic cover of a small can). Heat an iron wire or
            knitting-needle and burn through the plug along the
            axis to form a channel for the match. A top like this
Figure 23   will spin long and steadily.
               5. Figure 24 shows another top: a flat round box
            pierced by a sharpened match. For the box to fit tightly
            without sliding along the match, seal the hole with wax.
               6. A fancy top you see in Figure 25. Globular
            buttons with an eye are tied to the edge of a cardboard
Figure 24   disk with pieces of string. As the top rotates the
            buttons are thrown off radially, stretching the strings
            out taut and graphically demonstrating the action of
            our old friend, the centrifugal force.
               7. The same principle is demonstrated in another
            way by the top in Figure 26. Some pins are driven into
            the cork ring of the top with coloured beads threaded
            onto them so that beads can slide along the pin. As the
Figure 25
            top spins the beads are pushed away to the pin heads.
            If the spinning top is illuminated, the pins merge into
            a solid silvery belt with a coloured fringe of the merged
            beads. In order to enjoy the illusion spin the top on
            a smooth plate.
               8. A coloured top (Fig. 27). It is fairly laborious to
            make but the top will reward your efforts by
Figure 26
            demonstrating an astounding behaviour. Cut a piece of
            cardboard into a disk, make a hole at the centre to
            receive a pointed match. Clamp the match on either
            side of the disk with two cork disks. Now divide the
            cardboard disk into equal sectors by straight radial
            lines in the same way a round cake is shared out.
            54-55            For Young   Physicists

            Colour the sectors alternately in yellow and blue. What
            will you see as the top rotates? The disk will appear
Figure 27
            neither blue nor yellow, but green. The blue and yellow
            colours merge in your eye to give a new colour, green.
               You can continue your experiments on colour
            blending. Prepare a disk with sectors alternately
            coloured in blue and orange. Now, when the disk is
            spun it will be white, not yellow (actually it will be light
            grey, the lighter the purer the paints). In physics two
            colours that, when blended, give white are called
            "complementary". Consequently, our top has shown
            that blue and orange are complementary.
               If you have a good set of paints you can try an
            experiment that was first done 200 years ago by the
            great English scientist Isaac Newton. Paint the sectors
            of a disk with the seven colours of the rainbow which
            are: violet, indigo, blue, green, yellow, orange, and red.
            When all the seven colours are rotated together they
Figure 28   will produce a greyish-white. The experiment will help
            you to understand that the sunlight is composed of
            many colours.
               These experiments can be modified as follows: as the
            top spins throw a paper ring onto it and the disk will
            change its colour at once (Fig. 28).
               9. The writing top (Fig. 29). Make the top as just
Figure 29   described, the only difference being that its axle will
            now be a soft pencil, not match. Make the top spin on
            a cardboard sheet placed somewhat at an angle. The
            top will, as it spins, descend gradually down the
            inclined cardboard sheet, with the pencil drawing
            flourishes. These are easy to count and, since each one
            corresponds to,one turn of the top, by watching the top
            with a clock in hand* you can readily determine the
Figure 30   number of revolutions the top makes each second.
            Clearly, this would be impossible in any other way.
               A further form of the writing top is depicted in
            Fig. 30. Find a small lead disk and drill a hole at the
            centre (lead is soft and drilling it is easy), and a hole on
            either side of it.
               Through the centre hole a sharpened stick is passed,
            and through one of the side holes a piece of fishing-line

               * By the way, seconds can also be reckoned without a clock just
            by counting. To do so, you should at first drill yourself a bit to
            pronounce "one", "two", "three", etc., so that each number takes
            exactly one second to pronounce. Don't think that it's difficult, the
            practice shouldn't take more than 10 minutes.
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            (or bristle) is threaded so that it protrudes a bit lower
            than the end of the top axle. The fishing-line is fixed in
            with a piece of match. The third hole is left as it is, its
            only purpose is to balance the disk since otherwise the
            top won't spin smoothly.
               Our writing top is ready, but to experiment with it
            we need a sooty plate. Hold a plate over a smoky flame
            until it is covered with a uniform layer of dense soot.
            Then send the top spinning over the sooty surface. It
            will slide over the surface and the end of the fishing-line
            will draw, white on black, an intricate and rather
            attractive ornament.
               10. Our crowning effort is the last rig, the merry-go-
            round top. However, it is much easier to make that it
            might seem. The disk and stick here are just as in the

Figure 31

            earlier coloured top. Into the disk, pins with small flags
            are stuck symmetrically about the axis, and tiny paper
            riders are glued in-between the pins. Thus, you have
            a toy merry-go-round to amuze your younger brothers
            and sisters.

            When two boats, trams or croquet balls collide (an
            incident or move in a game) a physicist would call such
            an event just "impact". The impact lasts a split second,
            but if the objects involved are elastic, which is normally
            the case, then a lot happens in this instant. In each
            elastic impact physicists distinguish three phases. In the
            first phase both colliding objects compress each other
            at the place of contact. Then comes the second phase
            when the mutual compression reaches a maximum, the
            internal counteraction begins in response to the
            compression and prevents the bodies from compressing
            further, so balancing the thrusting force. In the third
            phase the counteraction, seeking to restore the body's
            shape deformed during the first phase, pushes the
            objects apart in opposite directions. The receding
            For Young   Physicists

            object, as it were, receives its impact back. In fact, when
            we observe, say, one croquet ball striking another,
            stationary, ball of the same mass, then the recoil makes
            the oncoming ball stop and the other ball roll forward
            with the velocity of the first.
               It is very interesting to observe a ball striking
            a number of other balls arranged in a file touching each
            other. The impact received by the first ball is, as it
            were, transferred through the file, but all the balls
            remain at rest and only the outermost one jumps away
            as it has no adjacent ball to impart the impact to and
            receive it back.
Figure 32

               This experiment can be carried out with croquet
            balls, but it is also a success with draughts or coins.
            Arrange the draughts in a straight line, it can be a very
            long one, but the essential condition is that they touch
            one another. Holding the first draught with a finger
            strike it on its edge with a wooden ruler, as shown.
            You will see the last draught jump away, with the rest
            of the draughts remaining in their places.

            An Egg in a Glass
            Circus conjurers sometimes surprise the public by
            jerking the cloth from a laid table so that everything-
            plates, glasses, and bottles - remain safely in place.
            This is no wonder or deceit, it is simply a matter of
            dexterity acquired by prolonged practice.
               Such a sleight-of-hand is too difficult for you to
Figure 33   attain but on a smaller scale a similar trick is no
            problem. Place a glass half-filled with water on a table
            and cover it with a postcard (or half of it). Further,
            borrow a man's wide ring and a hard-boiled egg. Put
            the ring on the top of the card, and stand the egg on
            the ring. It possible to jerk the card away so that the
            egg doesn't roll down onto the table?
              At first sight, it may seem as difficult as jerking the
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            table-cloth from under the table things. But a good
            snap with a finger on the edge of the card should do
            the trick. The card flies away and the egg... plunges
            with the ring safely into the water. The water cushions
            the blow and the shell remains intact.
              With some experience, you could try the trick with
            a raw egg.
              This small wonder is explained by the fact that
            during the fleeting moment of the impact the egg
            doesn't receive any observable speed but the postcard
            that was struck has time to slip out. Having lost its
            support, the egg drops into the glass.
              If the experiment is not at first a success, first
            practice an easier experiment in the same vein. Place
            half a postcard on the palm of your hand and a heavy
            coin on top of it. Now snap the card from under the
            coin. The card will fly away but the coin will stay.

            Unusual     Breakage
            Conjurers sometimes perform an elegant trick that
            seems amazing and unusual, though it can be easily
            explained. A longish stick is suspended on two paper
            rings. One of the rings is suspended from a razor blade,
            the other, from a clay pipe. The conjurer takes another
            stick and strikes the first one with all his strength.
Figure 34   What happens? The suspended stick breaks but the
            paper rings and the pipe remain absolutely intact!
               The trick can be accounted for in much the same
            way as the previous one. The impact was so fast that it
            allowed no time for the suspended stick's ends and the
            paper rings to move. Only the part of the stick that is
            directly subjected to the impact moves with the result
            that the stick breaks. The secret is thus that the impact
            was very fast and sharp. A slow, sluggish impact will
            not break the stick but will break the rings instead.
               The most adroit conjurers even contrive to break
            a stick supported by the edges of two thin glasses
            leaving the glasses intact.
               I do not tell you this, of course, to encourage you to
            do such tricks. You'll have to content yourself with
            a more modest form of them. Put two pencils on the
            edge of a low table or bench so that part of them
            overhang and place a thin, long stick on the over-
            hanging ends. A strong, sharp stroke with the edge of
            a ruler at the middle of the stick would break it in two,
            but the pencils would remain in their places.
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Figure 35

               Now it should be clear to you why it is difficult to
            crack a nut by the strong pressure of a palm, but the
            stroke of a fist does the j o b easily. When you hit it, the
            impact has no time to propagate along the flesh of your
            fist so that your soft muscles do not yield under the
            upthrust of the nut and act as a solid.
               For the same reason a bullet makes a small round
            hole in the window-pane, but a small stone traveling at
            a far slower speed breaks the pane. A slower push
            makes the window frame turn on its hinges, something
            neither the bullet nor the stone can make it do.
               Finally, one more example of the phenomenon is
            being able to cut a stem of grass by a stroke of a cane.
            By slowly moving the cane you can't cut a stem, you
            only bend it. By striking it with all your strength you
            will cut it, if, of course, the stem is not too thick. Here,
            as in our earlier cases, the cane moves too fast for the
            impact to be transferred to the whole of the stem. It
            will only concentrate in a small section that will bear
            all the consequences.

            Just Like a     Submarine
            A fresh egg will sink in water, a fact known to every
            experienced housewife. If she wants to find out whether
            an egg is fresh, she tests it in exactly this way. If an egg
            sinks, it is fresh; and if it floats, it is not suitable for
            eating. A physicist infers from this observation that
            a fresh egg is heavier than the same volume of fresh
            water. I say "fresh water" because impure (e.g. salt)
            water weighs more.
               It is possible to prepare such a strong solution of salt
            that an egg will be lighter than the amount of brine
            displaced by it. Then, following the principle of floating
            discovered in olden days by Archimedes, even the
            freshest of eggs will float in the solution.
            For Young   Physicists

               Use your knowledge in the following instructive
Figure 36   experiment. Try to make an egg neither sink nor float,
            but hang in the bulk of a liquid. A physicist would say
            that the egg is "suspended". You'll need a water
            solution of salt that is so strong that an egg submerged
            in it displaces exactly its own weight in the brine. The
            brine is prepared by the trial-and-error method: by
            pouring in some water if the egg surfaces and adding
            some stronger brine if it sinks. If you've got patience,
            you'll eventually end up with a brine in which the
            submerged egg neither surfaces nor sinks, but is at rest
            within the liquid.
               This state is characteristic of a submarine. It can stay
            under water without touching the ground only when it
            weighs exactly as much as the water it displaces. For
            this weight to be reached, submarines let water from
            the outside into a special container; when the
            submarine surfaces the water is pushed out.
               A dirigible-not an aeroplane but just a dirigi-
            ble - floats in the air for the very same reason: just like
            the egg in the brine it diplaces precisely as many tonnes
            of air as it weighs.

            Floating    Needle
            Is it possible to make a needle float on the surface of
            water like a straw? It would seem impossible: a solid
            piece of steel, although it's small, would be bound to
               Many people think this way and if you are among
            the many, the following experiment will make you
            change your mind.
               Get a conventional (but not too thick) sewing needle,
            smear it slightly with oil or fat and place it carefully on
            the surface of the water in a bowl, pail, or glass. To
            your surprise, the needle will not go down, but will stay
            on the surface.
               Why doesn't it sink, however? After all, steel is
            heavier than water? Certainly, it is seven to eight times
            as heavy as water, if it were under the water it wouldn't
            be able to surface like a match. But our needle doesn't
            submerge. To find a clue, look closely at the surface of
            the water near the floating needle. You'll see that near
            the needle the surface forms sort of a valley at the
            bottom of which lies our needle.
               The surface curvature is caused by the oil-smeared
            needle being not wetted by the water. You may have
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            noticed that when your hands are oily, water doesn't
            wet the skin. The feathers of water birds are always
            covered with oil exuded by a gland, which is why water
            doesn't wet feathers ("like water off a duck's back").
            And again this is the reason why without soap, which
            dissolves the oil film and removes it from the skin, you
            cannot wash your oily hands even by hot water. The
            needle with oil on it is not wetted by water either and
            lies at the bottom of a concavity supported by the
            water "film" created by surface tension. The film seeks
            to straighten and so pushes the needle out of the water,
            preventing it from sinking.
               As our hands are always somewhat oily, if you
            handle a needle it will be covered by a thin layer of oil.
Figure 37   Therefore, it is possible to make the needle float
            without specially covering it with oil-you'll only have
            to place it extremely carefully on some water. This can
            be made as follows: place the needle on a piece of
            tissue-paper, then gradually, by bending down the edges
            of the paper with another needle, submerge the paper.
            The paper will descend to the bottom and the needle
            will stay on the surface.
               Now if you came across a pondskater scuttling about
            the water surface, you won't be puzzled by it. You'll
            guess that the insect's legs are covered with oil and are
            not wetted by the water and that surface tension
            supports the insect on the surface.

            Diving bell
            This simple experiment will require a basin, but a deep,
            wide can would be more convenient. Besides, we'll need
            a tall glass (or a big goblet). This'll be our diving bell,
            and the basin with water will be our "sea" or "lake".
               There is hardly a simpler experiment. You just hold
            the glass upside down, push it down to the bottom of
            the basin holding it in your hand (for the water not to
Figure 38
            push it out). As you do so you'll see that the water
            doesn't find its way into the glass-the air doesn't let it
            in. To make the performance more dramatic, put
            something easily soaked, e.g. a lump of sugar, under
            your "bell". For this purpose, place a cork disk with
            a lump of sugar on it on the water and cover it by the
            glass. Now push the glass into the water. The sugar will
            appear to be below the water surface, but will remain
            dry, as the water doesn't get under the glass.
               You can perform the experiment with a glass funnel,
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            if you push it into the water, its wider end down and its
            narrow end covered tightly with a finger. The water
            again doesn't get inside the funnel, but once you
            remove your finger from the hole, thereby letting the
            air out of the funnel, the water will promptly rise into
            the funnel to reach the level of the surrounding water.
               You see that air is not "nothing", as some think, it
            occupies space and doesn't let in other things if it has
            nowhere to go.
               Besides, these experiments should           graphically
            illustrate the way in which people can stay and work
            under water in a diving bell or inside wide tubes that
            are known as "caissons". Water won't get into the bell,
            or caisson, for the same reason as it can't get into the
            glass in our experiment.

            Why Doesn't It Pour Out?
            The following experiment is one of the easiest to carry
            out, it was one of the first experiments I performed
            when I was a boy. Fill a glass with water, cover it with
            a postcard or a sheet of paper and, holding the card
            slightly with your fingers, turn the glass upside down.
            You can now take away your hand, the card won't
            drop and the water won't pour out if only the card is
            strictly horizontal.
               You can safely carry the glass about in this position,
            perhaps even more comfortably than usually since as
            the water won't spill over. As the occasion serves, you
            can astound your friends (if asked to bring some water
Figure 39   to drink) by bringing water in a glass upside down.
               What then keeps the card from falling, i.e. what
            overcomes the weight of the water column? The
            pressure of air! It exerts a force on the outside of the
            card that can be calculated to be much greater than the
            weight of the water, i.e. 200 grammes.
               The person who showed me the trick for the first
            time also drew my attention to the fact that the water
            must fill the glass completely for the trick to be
            a success. If it only occupies a part of the glass, the rest
            of the glass being filled by air, the trial may fail because
            the air inside the glass would press on the card
            balancing off the pressure of the outside air with the
            result that it might fall down.
               When I was told this, I set out at once to try it with
            a glass that wasn't fully filled in order to see for myself
            of the card would drop. Just imagine my astonishment
            54-55   For Young   Physicists

            when I saw that in that case too, it didn't fall! Having
            repeated the experiment several times, I made sure that
            the card held in place as securely as with the full glass.
               This has taught me a good lesson about how the
            facts of nature should be perceived. The highest
            authority in natural science must be experiment. Every
            theory, however plausible it might seem, must be tested
            by experiment. "Test and retest" was the motto of the
            early naturalists (Florentine academicians) in the 17th
            century, it is still true for 20th century physicists. And
            should a test of a theory indicate that experiment
            doesn't bear it out, one should dig for the clues to the
            failure of the theory.
                In our case we can easily find a weak point in the
            reasoning that once had seemed convincing. If we
            carefully turn back a corner of the card covering the
            overturned, partially filled glass, we'll see an air bubble
            come up through the water. What is it indicative of?
            Obviously the air in the glass was slightly rarefied,
            otherwise the outside air wouldn't rush into the space
            above the water. This explains the trick: although some
            air remained in the glass, it was slightly rarefied, and
            hence exerted less pressure. Clearly, when we turn the
            glass over, the water, as it goes down, forces some of
            the air out of the glass. The remaining air, which now
            fills up the same space, becomes rarefied and its
            pressure becomes weaker.
                You see that even simplest physical experiments,
            when treated attentively, can suggest fundamental ideas.
            These are those small things that teach us great ideas.

            Dry Out of Water
            You'll now see that the air surrounding us on all sides
            exerts a significant pressure on all the things exposed to
            it. The experiment I'm going to describe will show you
            more vividly the existence of what physicists call
            "atmospheric pressure".
                Place a coin (or metal button) on a flat plate and
Figure 40   pour some water over it. The coin will be under water.
            It's impossible, you are sure to think, to get it out from
            under the water with your bare hands without getting
            your fingers wet or removing the water from the plate.
            You're mistaken, it is possible.
                Proceed as follows. Set fire to piece of paper inside
            a glass and when the air has heated, upend the glass
            and put it on the plate near the coin. Now watch, you
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        won't have to wait long. Of course, the paper under the
        glass will burn out soon and the air inside the glass will
        begin to cool down. As it does so, the water will, as it
        were, be sucked in by the glass and before long it will
        be all there, exposing the plate's bottom.
           Wait a minute for the coin to dry and take it without
        wetting your fingers.
           The reason behind these phenomena is not difficult
        to understand. On heating, the air in the glass
        expanded, just as all bodies would do, and the extra
        amount of air came out of the glass. But when the
        remaining air began to cool down, its amount was no
        longer enough to exert its previous pressure, i.e. to
        balance out the external pressure of the atmosphere.
        Therefore, each square centimetre of the water under
        the glass was now subject to less pressure than the
        water in the exposed part of the plate and so no
        wonder it was forced under the glass by the extra
        pressure. In consequence, the water was not really
        "sucked in" by the glass, as it might seem, but pushed
        under the glass from the outside.
           Now that you know the explanation of the
        phenomenon in question, you will also understand that
        it is by no means necessary to use in the experiment
        a burning piece of paper or cotton wool soaked in
        alcohol (as is sometimes advised), or any flame in
        general. It suffices to rinse the glass with boiling water
        and the experiment will be as much of a success. The
        key thing here is to heat the air in the glass, no matter
        how that is done.
           The experiment can be performed simply in the
        following form. When you have finished your tea, pour
        a little tea into your saucer, turn your glass upside
        down while it is still hot, and stand it in the saucer and
        tea. In a minute or so the tea from the saucer will have
        gathered under the glass.

        Make a circle about a metre across out of a sheet of
        tissue-paper and then cut a circle a few centimetres
        wide in the middle. Tie strings to the edges of the large
        circle, passing them through small holes; tie the ends of
        the strings, which should be equally long, to a light
        weight. This completes the manufacture of a parachute,
        a scaled-down model of the huge umbrella that saves
        lives of airmen who, for some reason or other, are
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            compelled to escape from their aircraft.
               To test your miniature parachute in action drop it
Figure 41
            from a window in a high building, the weight down.
            The weight will pull on the strings, the paper circle will
            blossom out, and the parachute will fly down smoothly
            and land softly. This will occur in windless weather but
            on a windy day your parachute will be carried away
            however weak the wind and it will descend to the
            ground somewhere far from the starting point.
               The larger the "umbrella" of the parachute, the
            heavier the weight the parachute will carry (the weight
            is necessary for the parachute not to be overturned), the
            slower it will descend without a wind and the farther it
            will travel with a wind.
               But why should the parachute keep up in the air so
            long? Surely, you've guessed that the air stops the
            parachute from falling at once. If it were not for the
            paper sheet, the weight would hit the ground quickly.
            The paper sheet increases the surface of the falling
            object, yet adding almost nothing to its total weight.
            The larger the surface of an object, the more drag there
            is on it.
               If you've got it right, you'll understand why particles
            of dust are carried about by the air. It is widely
            believed that dust floats in air because it is lighter.
               What are particles of dust? Tiny pieces of stone, clay,
            metal, wood, coal, etc., etc. But all of these materials
            are hundreds and thousands of times heavier than air;
            stone, is 1,500 times heavier; iron, 6,000 times; wood,
            300 times, and so on.
               A speck of solid or liquid should infailingly fall down
            through the air, it "sinks" in it. It does fall, only falling
            it behaves like a parachute does. The point is that for
            small specks the surface-to-weight ratio is larger than
            for large bodies. Stated another way, the particles'
            surfaces are relatively large for their weight. If you were
            to compare a round piece of lead shot with a round
            bullet that is 1,000 times as heavy as the shot, the shot's
            surface is only 100 times smaller than the bullet's. This
            implies that the shot's surface per unit weight is 10
            times larger than the bullet's. Imagine that the shot
            shrinks until it becomes one million times lighter than
            the bullet, that is, turns into a speck of lead. Its
            "specific" surface would be 10,000 times larger than the
            bullet's. Accordingly, the air would hinder its motion
            10,000 times more strongly than it does the bullet's.
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            That's why it would hover in the air hardly falling and
            being carried by the slightest wind away and even

            A Snake and a        Butterfly
            Cut a circle about the size of a glass hole from
            a postcard or a sheet of strong paper. Cut a spiral in it
Figure 42   in the form of a coiled-up snake, as shown in Fig. 42.
            Make a small recess in the tail to receive a knitting-pin
            fixed upright. The coils of the snake will hang down
            forming sort of a spiral stairs.
                N o w that the snake is ready, we can set out to
            experiment with it. Place it near a hot kitchen stove:
            the snake will spin and the faster, the hotter the stove.
            Near any hot object (a lamp or tea kettle, etc.) the
            snake will rotate while the object remains hot. So, the
            snake will spin very fast if suspended above a kerosene
            lamp from a piece of string.
                What makes the snake rotate? The same thing that
            makes the arms of a windmill r o t a t e - t h e flow of air.
            Near every heated object, there is an air flow moving
Figure 43   upwards. This flow occurs because air, just like any
            other material, expands on heating and becomes
            thinner, i. e. lighter. The surrounding air, which is
            colder and thus denser and heavier, displaces the hotter
            air, making it rise, and occupies its place. But the fresh
            portion of air heats at once and, just like the first one is
            ousted by a yet fresher amount of colder air. In this
            way, each heated object gives rise to an ascending flow
             of air around it, which is maintained all the time the
             object is warmer than the surrounding air. In other
             words, a barely noticeable warm wind blows upwards
             from every heated object. It strikes the coils of our
             paper snake making it rotate, jyst as wind makes the
             arms of a windmill rotate.
                Instead of a shake you can use a piece of paper in
             another shape, for example a butterfly. Cut it out of
             tissue-paper, bind in the middle and suspend from
             a piece of a very thin string or hair. Hang the butterfly
             above a lamp and it will rotate like a live one. Also, the
             butterfly will cast its shadow on the ceiling and the
             shadow will repeat the motions of the rotating paper
             butterfly magnified up. It'll seem to an uninitiated
             person that a large black butterfly has flown into the
             room and is hectically hovers under the ceiling.
                You can also make as follows: strick a needle into
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            a cork and place the paper butterfly on the needle's tip
            at the point of equilibrium which can be found by trial
            and error. The butterfly will rotate quickly if placed
            above a warm thing, in fact putting your palm under it
            will be enough for the butterfly to rotate.
               We come across the expansion of air as it heats and
            ascending warm currents everywhere.
               It is well known that the air in a heated room is the
            warmest near the ceiling and the coldest near the floor.
            That's way it seems sometimes that there is a draught
            near our feet when the room hasn't properly heated up.
            If you leave the door from a warm room to a colder
            one ajar, cold air will flow into the warm one along the
            floor and warm air will flow out along the ceiling. The
            flame of candle placed near the door will indicate the
            direction of these flows. If you want to keep the
            warmth in a heated room you should see to it that no
            cold air comes in from under the door. You need only
            to cover the gap by a rug or just a newspaper. Then the
            warm air won't be ousted from below by the colder one
            and won't leave the room through holes higher up in
            the room.
               And what is the draught in a furnace or a chimney
            stack but an ascending flow of warm air?
               We could also discuss the warm and cold flows in
            the atmosphere, trade winds, monsoons, breezes and
            the like but it would lead us too far astray.

            Ice in a Bottle
            Is it easy to get a bottle full of ice during the winter? It
            would seem that nothing could be easier when it is
            frosty outdoors. Just put a bottle of water outside the
            window and let the frost do the job. The frost will cool
            the water and you will have a bottleful of ice.
               But if you actually try to do the experiment, you'll
            see that it is not that easy. You will obtain ice but the
            bottle will be destroyed in the process, it'll burst under
Figure 44   the pressure of the freezing water. This occurs because
            water, on freezing, expands markedly, by about a tenth
            of its volume. The expansion is so powerful that it
            bursts both a corked bottle and the bottleneck of an
            open bottle, the water frozen in the neck becomes, as it
            were, an ice cork.
               The expansion of freezing water can even break
            metal walls if they are not too thick. So, water can
            break the 5-cm walls of a steel bomb. N o wonder that
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            water pipes burst so often in winter.
               The expansion of water on freezing also accounts for
            the fact that ice floats on water and doesn't sink. If
            water contracted on cooling, just like all other liquids
            do, then ice wouldn't float on the water's surface but
            would go down to the bottom. And those of us in
            northern countries wouldn't enjoy skating and
            travelling on the ice of our rivers and lakes.

            To Cut Ice and... Leave It One Piece
            You may have heard that pieces of ice "freeze up"
            under pressure. This doesn't mean that pieces of ice
            freeze up more strongly when exposed to pressure. On
            the contrary, under strong pressure ice melts, but once
            the cold water produced in the process is free of the
            pressure, it refreezes (as its temperature is below 0°C).
            When we compress pieces of ice, the following occurs.
            The ends of the parts that contact each other and are
            subject to high pressure melt, yielding water at
            a temperature below zero. This water fills in tiny
            interstices between the parts that are sticking out and
            when it is not subjected to the high pressure any more
            it freezes at once, thus soldering the pieces of ice into
            a solid block.
               You can test this by an elegant experiment. Get
            a beam of ice, and support its ends by the edges of two
            stools, chairs or the like. Make a loop of a thin steel
            wire 80 centimetres long and put it round the beam, the
            wire should be 0.5 millimetre or a little less thick.
            Finally,    suspend     something    heavy     (about    10
            kilogrammes) from the ends of the wire. Under the
            pressure of the heavy object the wire will bite into the
            ice and cut slowly through the whole of the beam but...
            the beam will still remain one piece. You may safely
            take it in your hands as it will be intact as if it had not
            been cut!
               After you've learned about the freezing up of ice,
            you'll see why it works. Under the thrust of the wire
Figure 45   the ice melted but the water flowed over the wire and
            free of the pressure refroze at once. In plain English,
            while the wire cut the lower layers, the upper layers
            were freezing again.
               Ice is the only material in nature with which you can
            do this experiment. It's for this reason that we can
            skate and toboggan over ice. When a skater presses all
            his weight on his skates, the ice melts under the
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pressure (if the frost is not too severe) and the skate
slides along where it again melts some ice and the
process occurs continuously. Wherever the skate goes
a thin layer of ice turns into water that when free of the
pressure refreezes. Therefore, although the temperature
might be below freezing point, the ice is always
"lubricated" with water under skates. That's why it's so

Sound   Transmission
You may have observed from a distance a man using
an axe or a carpenter driving in nails. You may then
have noticed an unusual thing, you do not hear the
stroke when the axe touches the tree or when the
hammer hits a nail, but you hear it later when the axe
(or hammer) is ready for the next stroke.
   Next time you happen to observe something similar,
move a little forward or backward. After trying several
times you'll find a place where the sound of a stroke
comes just at the moment of a visible stroke. Then
return to where you started and you'll again notice the
lack of coincidence between the sound and the visible
   Now it should be easy for you to guess the reason
behind this enigmatic phenomenon. Sound takes some
time to cover the distance from the place where it
originated to your ear; on the other hand, light does it
nearly instantaneously. And it may happen that while
the sound is travelling through the air to your ear, the
axe (or hammer) will have been raised for a new stroke.
Then your eyes will see what your ears hear and it'll
seem to you that the sound comes when the tool is up
and not when the tool is down. But if you move in
either direction just a distance, covered by the sound
during a swing of the axe, then by the time the sound
reaches your ear the axe will strike again. Now, of
course, you'll see and hear a stroke simultaneously,
only it'll be different strokes since you'll hear an earlier
stroke, perhaps the last but one or even earlier.
   What is the distance covered by sound in one
second? It has been measured exactly, but approxi-
mately it is about 1/3 of a kilometre. So sound covers
one kilometre in 3 seconds, and if the wood-cutter
swings his axe twice a second, you'll have to be 160
metres away for the sound to coincide with the axe as
he raises it. But light travels each second in air almost
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            a million times as far as sound. So you can understand
            that for any distances on earth we can safely take the
            speed of light to be infinite.
               Sound is transmitted not only through the air but
            also through other gases, liquids and solids. So in
            water, sound travels four times faster than in air, and
            under water sound can be heard distinctly. People
            working in underwater caissons can hear sounds from
            the shore perfectly and anglers will tell you how fish
            scatter at the slightest suspicious noise from the shore.
               Elastic solids are better still as sound transmitters,
            e. g. cast iron, wood, and bone. Put your ear to the end
            face of a long wood beam or a block and ask
            somebody to tap it slightly at the other end. You'll hear
            the dull sound of the stroke transmitted through the
            entire length of the beam. If it's rather quiet and
            spurious noises don't interfere, you can even hear
            a clock ticking at the opposite end of your beam.
            Sound is transmitted equally well along iron rails or
            beams, cast iron tubes, and even soil. If you put your
            ear to the ground you can hear the clatter of horses'
            hoofs long before the sound comes through the air and
            in this way you can hear thunder that is so far away
            that no sound comes to you by air at all.
               Only elastic solids transmit sound so well, soft tissues
            and loose, inelastic materials are very poor sound
            transmitters since these "absorb" it. That's why they
            hang thick curtains near doors if they don't want any
            sound to reach an adjacent room. Carpets, soft furni-
            ture and clothes have the same effect on sound.

            A Bell
            Among the materials distinguished for their perfect
            sound transmission I've mentioned bones. Do you want
            to make sure that the bones of your skull have this
            property? Hold the ring of a pocket watch with your
            teeth and close your ears with your hands. You'll still
            quite distinctly "hear" the measured strokes of the
Figure 46   balance, and they'll be louder than the ticking perceived
            through the ear. This sound comes to your ears
            through the bones of your skull.
               A further fascinating experiment testifying to the
            good transmission of sound through your skull. Tie
            a soup spoon in the middle of a piece of string so that
            the string has two loose ends. Press these ends with
            your fingers to your closed ears and, leaning forward
            74-75   For Young   Physicists

            for the spoon to swing freely, make it strike something
            solid. You'll hear a low-pitched drone as if a bell is
            ringing near your ears.
               The experiment comes out better if you use
            something heavier instead of the spoon.

            A Frightening       Shadow
            One evening my brother Alex asked, "Want to see
            something unusual? Come into this room."
               The room was dark. Alex took a candle and we
            walked in. I led the way and so was the first to enter
            the room. But suddenly I was stunned: an incongruous
            monster eyed me from the wall. Flat as a shadow, it
            stared at me.
               To tell the truth, I got a little frightened. I might
            have taken to my heels had it not been for my brother's
            laughter behind me.
               I turned round and saw the reason. There was

Figure 47

            a mirror on the wall covered with a sheet of paper that
            had eyes, a nose, and a mouth cut in it. Alex had so
            directed the candle's light that these parts of the mirror
            reflected directly onto my shadow.
               Thus, I was scared by my own shadow.
               But later, when I attempted to play this joke on my
            friends, it turned out that arranging the mirror properly
            is not that easy. It took a lot of practice to master the
            art. Light rays are reflected in a mirror according to the
            following rule: the angle at which light rays strike the
            mirror equals the angle at which they are reflected.
            After I'd learned the rule it was no problem to work
            out where to locate the candle with respect to the
            mirror for the light spots to be cast at the required
            place on the shadow.
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            To Measure Light          Brightness
            At twice the distance, clearly, a candle illuminates much
            weaker. But how many times weaker? Two? No, if you
            place two candles at double the distance, you won't
            obtain the previous illumination. In order to obtain the
            earlier illumination at double the distance you'd have
            to put two times two, i.e. four candles, not two. At
            triple the distance you'd need three times three, i.e.
            nine candles, not three, and so forth. It follows that at
            twice the distance illumination is four times weaker; at
            three times the distance, nine times weaker; at four
            times the distance, 16 times; and at five times the
            distance, 5 x 5 or 25 times weaker, and so on. This is
            the law of weakening illumination with distance. Note
            in passing that this is also the law of sound attenuation
            with distance. For example, sound attenuates on six
            times the original distance by 36 times, not by 6 times*.
               Knowing this law we can make use of it to compare
            the brightness of two lamps, or any two light sources in
            general. For instance, you wish to compare the
            brightness of your lamp with that of a conventional
            candle, in other words, you want to find out how many
            candles you need to replace the lamp to obtain the
            same illumination.
               For this purpose place the lamp and a burning
            candle at one end of a table and at the other you stand
            a sheet of white cardboard clamped between books as

Figure 48

                * This explains why a whisper from your neighbour drowns the
            loud voice of an actor on the stage in a theatre. If the actor is only
            10 times farther away from you than your neighbour, then the
            actor's voice is attenuated 100 times more than what you'd hear if
            the same sound came to you from the lips of your neighbour. It's
            not surprising then that for you the actor's voice is weaker than the
            whisper. For exactly the same reason it's important for students to
            keep quiet when the teacher speaks. The teacher's words reaching
            students (especially those far away) are so attenuated that even a soft
            whisper from a neighbour will muffle them completely.
            74-75   For Young   Physicists

            shown. Just in front of the sheet fix up a stick, e.g.
            a pencil, also upright. It will cast two shadows onto the
            cardboard, one from the lamp and the other from the
            candle. The density of the two shadows is, generally
            speaking, different because both are lit, one by the
            bright lamp, the other by the dimmer candle. By
            bringing the candle nearer you can achieve a situation
            in which both shadows will have the same "blackness".
            This will mean that the illumination due to the lamp
            just equals that due to the candle. But the lamp is far-
            ther away from the cardboard than the candle.
            Measuring how many times farther away will tell you
            how many times the lamp is brighter than the candle.
            If, say, the lamp is three times farther away from the
            cardboard than the candle, then its brightness is 3 x 3,
            i.e. nine times the brightness of the candle. Remember
            the law?
               Another way of comparing the luminous intensity of
            two sources relies on the use of an oil spot on a sheet
            of paper. The spot will seem light if illuminated from
            behind, and dark if lit from the front. So the sources to
            be compared can be placed at such distances that the
            spot will seem equally illuminated on either side. Then
            it only remains to measure the respective distances and
            repeat the previous process. And in order that both
            sides of the spot might be compared it is a good idea to
            place the paper near a mirror, you should know how.

            Upside Down
            If your home has a room facing south, you could easily
            make it into a physical device that has an old Latin
            name camera obscura. You'll have to close the window
            with a shield, made of a plywood or cardboard glued
            with dark paper with a small hole made in it. On a fine
            sunny day close the doors and windows to darken the
            room and place a large sheet of paper or a sheet
            opposite the hole. This will be your "screen". You'll
            immediately see on it a reduced image of what can be
Figure 49
76-77   For Young   Physicists

        seen from the room through the hole. Houses, trees,
        animals, and people, everything will appear on the
        screen in its natural colours, but... upside down.
           What does this experiment prove? That light
        propagates in straight lines. The rays from the upper
        and lower parts of an object cross in the hole and
        travel on so that now the top rays appear below and
        the bottom rays above. If the rays were not straight but
        curved or broken, you'd have got something different.
           Significantly, the shape of the hole has no effect
        whatsoever on the image. You might drill a round hole,
        or make a square, triangular, hexagonal, or other
        h o l e - t h e image on the screen would be the same. Did
        you happen to observe oval light circles under a dense
        tree? These are nothing but images of the Sun painted
        by the rays that pass through various gaps between the
        leaves. The images are roundish because the sun is
        round, and elongated because the rays are obliquely
        incident on the ground. Put a sheet of paper at right
        angles to the solar rays and you'll obtain round spots
        on it. During solar eclipses when the dark sphere of the
        moon blots out the sun leaving only a bright crescent,
        the small spots under trees turn into small crescents as
           The old photographer's camera, too, is nothing but
        a camera obscura, the only difference being that at the
        hole an objective lens is fitted for the image to be
        brighter and clearer. The back wall is a frosted glass on
        which the image is produced, upside down of course.
        The photographer can only view it if he covers himself
        and the camera by a dark cloth to keep out any
        spurious light.
           You can make a simple model of this sort of camera.
        Find a closed elongated box and drill a hole through
        one wall. Remove the wall opposite the hole and
        stretch over the gap an oiled piece of paper instead-a
        substitute for the frosted glass. Bring the box into our
        dark room and place it so that its hole is just opposite
        the hole in the darkened window. On the back side
        you'll see a distinct image of the outside, again upside
        down of course.
           Your camera is convenient in that you no longer
        need a dark room and you can bring it out into the
        open and put it where it suits you. You'll only need to
        cover your head and the camera with a dark cloth for
        the spurious light not to interfere.
            For Young   Physicists

            Overturned Pin
            We have just discussed the camera obscura and a way
            of manufacturing it but we omitted one interesting
            thing: every human being always has a pair of small
            cameras like that about him or her. These are our eyes.
            Just fancy, your eye is just like the box that you were
            shown how to make above. What we call the pupil of
            the eye is not a black circle on the eye but a hole
Figure 50   leading into the inside of your organ of sight. The hole
            is covered with a transparent envelope on the outside
            and with a jelly-like and transparent substance
            underneath. Next to the pupil behind it is the
            crystalline lens having the form of a convexo-convex
            glass, and the inner cavity of the eye between the
            crystalline lens and the back wall, on which the image
            is produced, is filled with a transparent substance.
            A cross section through the eye is given in Fig. 50.
            Despite all these distinctions the eye is still a camera
            obscura, only an improved one, as the eye produces
            high-quality, distinct images. The images at the back of
            the eye are minute. So, an 8-m high lamp-post seen 20
            metres away from the eye is only a tiny line, about
            5 millimetres long, at the back of the eye.
                But the most interesting thing here is that although
            all the images are upside down, we perceive them as
            they are. This turning over is due to long habit. We are
            used to seeing with our eyes so that each visual image
            obtained is converted into its natural position.
                That this is really true, you could test by an
            experiment. We'll attempt to contrive it so that we get
            at the back of the eye not an inverted, but direct image
            of an object. What will we see then? Since we are used
            to inverting every visual image, we'll invert this one as
            well. Accordingly, we'll in this case too see an inverted
            image, not a direct one. In actual fact that is exactly
            what happens and the following experiment will
            demonstrate it in a fairly graphic manner.
Figure 51
                Make a pinhole in a postcard and hold it against
            a window or a lamp about 10 centimetres away from
            your right eye. Hold the pin between you and the
            postcard so that its head is opposite the hole. With this
            arrangement you'll see the pin as if it were behind the
            hole, and what is of the more importance here, upside
            down. This unusual situation is presented in Fig. 51.
            Move the pin to the right and your eyes will tell you
            it's moved to the left.
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            The explanation is that the pin at the back of your
        eye is here depicted not upside down but directly. The
        hole in the card plays the role of a light source
        producing the shadow of the pin. The shadow falls on
        the pupil and its image is not inverted as it's too close
        to the pupil. On the back wall of the eye a light spot is
        p r o d u c e d - t h e image of the hole in the card. On it the
        dark silhouette of the pin is seen which is its shadow,
        the right way up. But it seems to us that through the
        hole in the card we see the pin behind the card (as only
        the part of the pin that gets in the hole is seen) and
        inverted at that because our eyes are in the habit of
        turning images upside down.

        Igniting with Ice
        When a boy, I liked watching my brother lighting
        a cigarette with a magnifying glass. He would put the
        glass in the sunlight and train the spot of light on the
        cigarette end. After a while it would begin to give off
        a bluish smoke and smolder.
           One winter day Alex said, "You know, it's possible to
        light a cigarette with ice, too."
           "With ice?"
           "Ice doesn't ignite it, of course, the sun does. Ice only
        collects solar rays, just like this glass."
           "So you want to make a magnifying glass out of
           "I can't make glass of ice, nobody can, but we could
        make a burning lens from ice."
           "What's a lens?"
           "We'll shape a piece of ice like this glass and it will
        be a lens: round and convex which means thick in the
        middle and thin at the edges."
           "And will it ignite things?"
           "But it's cold!"
           "What of it? Let's try."
           To begin with, my brother told me to fetch
        a washing basin. When I did he rejected it: "Nothing
        doing. You see, the bottom is flat. We need a curved
           When I brought a suitable basin, Alex poured some
        clean water into it and put it outside, the temperature
        outdoors being below freezing point.
           "Let it freeze down to the bottom. We'll then have an
        ice lens with one side flat and the other convex."
            74-75   For Young   Physicists

               "So big?"
               "The bigger, the better, it'll catch more sunlight."
               First thing in the morning I ran to inspect our basin.
            The water had frozen right through to the bottom.
               "What a good lens we'll have," Alex said tapping the
            ice with finger. "Let's take it out of the basin."
               This turned out to be no problem. Alex put the icy
            basin into another one containing hot water and the ice
            at the walls melted quickly. We got the ice basin out
            into the yard and placed the lens on a board.
Figure 52      "Good weather, ins't it!" Alex screwed up his eyes in
            the sunlight, "Ideal for igniting. Just hold the cigarette."
               I did so and my brother, taking hold of the lens with
            both hands turned it towards the sun but so that he
            wasn't in the way of the rays himself. He took aim
Figure 53
            painstakingly but eventually succeeded in training the
            lightspot directly on the end of the cigarette. When the
            spot rested on my hand, I felt it was hot and already
            I had no doubt that the ice would light the cigarette.
               Indeed, when the spot got onto the end of the
            cigarette and had stayed there for about a minute, the
            tobacco smoldered and discharged some bluish smoke.
               My brother took a puff at the cigarette, "Here you
            are, we've lit it with ice. In this way you could make
            a fire without matches even at the pole, if only you had

            Magnetic     Needle
            You can already make a needle float on the surface of
            water. Here you'll have to use your skill in a new and
            more impressive experiment. Find a magnet, if only
            a small horse-shoe one. If you bring it near the saucer
            with a needle floating in it, the needle will obediently
            approach the appropriate edge of the saucer. The effect
            will be more noticeable, if before placing the needle on
            the water you pass the magnet several times along it
            (but only use one end of the magnet in one direction
            only). This turns the needle itself into a magnet, there-
            fore it'll even approach a nonmagnetized iron object.
              You can make many curious observations with the
            magnetic needle. Leave it alone without attracting it by
            a piece of iron or the magnet and it'll orient itself in the
            water in one direction, namely north-south, just like the
            needle of a compass. Turn the saucer and the needle
            will still point to the north with one end and to the
            south with the other. Bring one end (pole) of the
      74-75   For Young   Physicists

              magnet to an end of the needle and you'll find that it
              won't be attracted to the magnet at that end. It may
              turn away from the magnet in order that the opposite
              end might approach. This is a case of an interaction
              between two magnets. The law of this interaction states
              that unlike ends (the north pole of one magnet and the
Figure 54
              south pole of another) are attracted and like ones (both
              north or south) are repelled.
                 Having investigated the behaviour of the magnetized
              needle, make a toy paper boat and hide your needle in
              its folds. You might astonish your uninitiated friends by
              controlling the motion of the boat without so much as
              touching it: it would obey every motion of your hand.
              Of course, you would be holding the magnet so that
              the spectators wouldn't suspect it.

              Magnetic     Theatre
              Or rather circus, as starring in it are rope dancers cut
              out of paper (of course).
                 First of all, you have to make the circus building out
              of cardboard. At the bottom of it you'll stretch a wire
              and fix above the stage a horse-shoe magnet, as shown.
                 Now to the artists. They are cut out of paper, their
              stance being chosen to suit the purpose. The only
              condition is that their height be equal to the length of
              a needle glued on from behind along the length of the
Figure 55

              figure. You could use two or three drops of sealing-wax
              for the glue.
                 If a figure like this is installed onto the "rope", it not
              only won't fall, but will stay upright pulled by the
              magnet. By slightly jerking the wire you'll animate your
              rope dancers. They'll swing and jump all the while
              keeping their balance.
6 — 975
            74-75   For Young   Physicists

            Electrified   Comb
            Even if you're ignorant of electricity and not even
            acquainted with its ABC, you can still do a number of
            electrical experiments that would be fascinating and
            will, in any case, be useful when you meet this amazing
            force of nature in future.
               The best place for these electrical experiments is
            a warm room in a frosty winter. These experiments are
            especially successful in dry air, and in winter warm air
            is far drier than air at the same temperature during
               Now, to our experiments. You may have passed
            a conventional comb over dry (completely dry) hair. If
            you did so in a warm room in full silence, you may
            have heard some slight crackling on the comb. Your
            comb had been electrified by friction with the hair.
               The comb can also be electrified by material other
            than hair. If you rub it against a dry woollen fabric (a
            piece of flannel, say) it also acquires electrical
            properties and to quite a larger degree. These
            properties manifest themselves in a wide variety of
            ways, notably by attracting light objects. Bring
Figure 56
            a rubbed comb close to some pieces of paper, chaff,
            a ball of elder core, etc. and these small things will all
            stick to the comb. Make tiny ships of light paper and
            launch them on water. You'll be able to control the
            movements of your paper fleet using an electrified
            comb like a magic wand. You could stage the
            experiment in a more impressive way. Place an egg in
            a dry egg-support and balance a rather long ruler on it.
             As the electrified comb approaches one of its ends the
             ruler will turn fairly quickly. You can make it follow
             the comb obediently moving it in any direction and
             making it rotate.

            An Obedient Egg
            Electrical behaviour is inherent not only in the comb
            but in other things as well. A rod of sealing-wax rubbed
            against a piece of flannel or the sleeve of your coat, if
            it's woollen, behaves in the same way. A glass rod or
            tube, too, is electrified if rubbed by silk. But the
            experiment with silk is only a success in exceedingly
            dry air and only then if both the silk and the glass are
            well dried by heating.
                Here is a further funny experiment on electrical
            attraction. Empty a chicken egg through a small hole,
82-83       For Young     Physicists

            which is best done by blowing the contents out through
            another hole at the opposite end. You've thus obtained
            an empty shell (the holes are sealed with wax). Put it
            on a smooth table, board or large plate and, using the
            electrified rod, make the empty egg roll obediently after
            it. An outsider, not aware that the egg is empty, would
Figure 57

            be bewildered by experiment (invented by the English
            scientist Faraday). A paper ring or a light ball, too,
            follow an electrified rod.

            Mechanics teaches that one-sided attraction, or any
            one-sided action, in general, doesn't exist. Any action is,
            in fact, an interaction. In consequence, if the electrified
            rod attracts various things, then it itself is attracted to
            them. To bear this out you have only to make the
Figure 58
            comb, or rod, easily movable, e.g. by suspending it
            from a loop made of a piece of thread (the thread
            should preferably be a silk one).
              Then, you will quickly find that any electrified thing-
            your hand, say-attracts the comb making it turn, and
            so forth.
               To repeat, this is a general law of nature. It shows up
            always and everywhere-any action is an interaction of
            two bodies affecting each other in opposite directions.
            Nature doesn't know of an action that is one-sided and
            doesn't involve the interaction of another body.

            Electrical     Repulsion
            Let's return to the experiment with the suspended
            electrified comb. We've seen that it is attracted by any
            electrified body. It would be of interest to test the way
            in which another, also electrified, thing affects it. An
            experiment will convince you that two electrified bodies
            can interact in different ways. If you bring an electrified
            glass rod to the electrified comb, the two things will
            attract each other. But if you bring an electrified
            sealing-wax rod or another comb to the comb, the
            interaction will be repulsive.
               The physical law describing this fact of nature states:
            unlike charges attract, like charges repel. Like charges

            74-75   For Young   Physicists

            will be those on plastics and sealing-wax (the so-called
            amber or negative, charge) and unlike charges are those
            on amber and glass which is positive. The ancient
            names "amber" and "glass" charges have now gone out
            to use, being completely replaced by the names
            "negative" and "positive" charges.
               The repulsion of like-charged things lies at the basis
            of a simple device to detect electricity-the so-called
            electroscope. The word "scope" comes from Greek and
            means to "indicate", it enters words like "telescope",
Figure 59   "microscope", and so forth.
               You can make this simple device on your own.
            Through the middle of a cardboard circle or a cork
            that fits the neck of a jar or bottle, a rod is passed, part
            of it protruding from the top. To the end of the rod
            two strips of foil or tissue-paper are attached using
            wax. Next the neck is plugged with the cork or
            cardboard circle, sealing the edges with sealing wax.
            The electroscope is ready to use. If now you bring an
            electrified thing to the protruding end of the rod, the
            two strips will become electrified, too. They charge
            up simultaneously and, therefore, separate due to
            electrostatic repulsion. The separation of the strips is
Figure 60   the indication that the thing that touched the
            electroscope rod is electrified.
               If you are no good at handiwork, you could make
            a simpler version of the device. It won't be as
            convenient and sensitive, but will still work. Suspend
            two elder-core balls on a stick from pieces of string so
            that they hang in contact with each other. That's all.
            O n touching a ball with a thing being tested you'll
            notice that the other ball deflects if the thing is charged.
               Finally, in the accompanying figure you can see yet
            another form of a primitive electroscope. A foil strip,
            folded in two, is suspended from a pin stuck into
            a cork. Touching the pin with an electrified thing
            makes the strips separate.

            One Characteristic       of      Electricity
            With the help of an easily manufactured makeshift
            device you can observe the interesting and very
            important feature of electricity - to accumulate on the
            surface of an object only, and on protruding parts at
              Cement a match vertically to a match box using
            a sealing-wax drop, then make another such support.
84-85       74-75   For Young   Physicists

            Now cut out a paper strip about a match-length wide
Figure 61   and three match-lengths long. Turn the ends of the
            paper strip into a tube so that you could fix it to the
            supports. Glue three or four narrow ribbons of thin
            paper-tissue to the either side of the strip (Fig. 61) and
            fix the assembly on the matches.
               O u r device is ready for experiments. Touch the
            straight strip with an electrified sealing-wax rod and the
            paper and all the ribbons on it will charge up
            simultaneously. This can be judged by the ribbons
            sticking out on either side. Now arrange the supports
            so that the strip curves into an arc, and charge it up.
            The strips will now stick out on the convex side only,
            those on the concave one will dangle as before. What
            does this indicate? That the electric charge has only
            accumulated on the convex side. If you make the strip
            into an S-shape, you'll see that the electric charge is
            only present on the convex parts of the paper.
A Sheet of Newspaper

What is to "Look with Your Mind"? • Heavy   Newspaper

"Agreed. This evening we are performing electrical
experiments," my brother proclaimed tapping the tiles
of the warm stove.
   I was delighted, "Experiments? New experiments!
When? Right now? I'd like to now!"
   "Patience, my friend. The experiments will be this
evening. Now I must be off."
   "To get the machine?"
   "What machine?"
   "Electric. We'll need a machine for our experiments."
   "The machine that we'll need is already available, it's
in my bag... And don't you dare delve in there while
I'm away," Alex had read my thoughts. He went on to
say, putting his coat on, "You'll find nothing, and will
only make a mess."
   "But the machine is there?"
   "There, don't worry."
   My brother went out, carelessly leaving the bag with
the machine in on a small table in the hall.
   If iron could feel, it would feel near a magnet exactly
what I was feeling left alone with my brother's bag. The
bag was pulling me, attracting all my feelings and
thoughts. It was absolutely impossible to think about
something different or divert my eyes from the b a g -
   It's so strange that an electric machine can go inside
a bag. I did not imagine it to be that flat. The bag
wasn't locked and I carefully peaped inside... Something
wrapped in newspaper. A small box? No, books.
Books? Only books, nothing more in the bag? I should
have understood at once Alex was joking: how can you
possibly hide an electric machine in a bag!
   Alex came back with empty hands and guessed at
once the reason for my sorrowful looks.
   "We seem to have visited the bag don't we?" he said.
   "Where is the machine?" I answered with a question.
   "In the bag, didn't you see?"
   "There are only books in there."
   "And the machine! You didn't look very far. What
did you look with?"
   "With what? Why, with my eyes?"
   "That's just it, with your eyes. You didn't use your
brains. It's not enough just to look, you have to
understand what you see. That is called looking with
your mind."
            A Sheet of   Newspaper

                "How do you look with your mind?"
Figure 62      "Do you want me to show to you the difference
            between looking with your eyes and looking with the
            whole of your head?"
               My brother produced a pencil and drew a figure
            (Fig. 62) on a sheet of paper.
               "The double lines here are railways, the single
            ones - highways. Take a look and say which of the
            railways is longer, the one from 7 to 2 or from 1 to 3?"
               "From 1 to 3, of course."
               "You see it with your eyes. But now look at the
            figure with the whole of your head."
               "But how? I can't."
               "Like this. Imagine that a straight line is drawn from
            J at a right angle to the lower highway 2-5," my
            brother drew a dash line in his drawing. "How will my
            line separate the highways? Into what parts?"
               "In two."
               "Exactly. This implies that all the points of this dash
            line are equidistant from 2 and 3. What will you say
            now about point 1 ? Is it closer to 2 or J ?"
               "Now I see that it's the same distance from 2 and 3.
            But earlier it seemed that the right-hand railway was
            longer than the left."
               "Earlier you only looked with your eyes, but now
            you're using your head. See the difference?"
               "I see. But where's the machine?"
               "What machine? Oh, yes, the electric machine. In the
            bag. It's still there. You didn't notice because you didn't
            look with your mind."
               My brother took a bundle of books out of the bag,
            carefully unwrapped it from a large newspaper sheet
            and gave it to me.
               "Here's our electric machine."
               I looked at the newspaper in bewilderment.
               "Do you think it's only paper and nothing more?"
            My brother went on to say, "According to your eyes,
            yes. But someone who can use his brains will perceive
            a physical device in this paper."
               "Physical device? To make experiments?"
               "Yes. Hold the newspaper in your hands. It's light,
            isn't it? And, of course, you'll think that you can
            always lift it even with a single finger. But now you'll
            see that this very newspaper can at times be very heavy.
            Give me that ruler."
               "It's got serrated and isn't good for anything."
               "All the better, it doesn't matter if it gets broken."
            86-87   A Sheet of   Newspaper

               Alex put the ruler on a table so that a part of it
            overhung the edge.
               "Touch the protruding end. It's easy to press it down,
            isn't it? Well, try to press it down after I've covered the
            other end with the newspaper."
               He spread the newspaper on the table over the ruler,
            carefully smoothing tne folds.
               "Now take a stick and strike the protruding part of
            the ruler very hard. Strike with all you strength!"
               I swung the stick back, and said, "I'll strike it so
            hard that the ruler will break through the paper and hit
            the ceiling!"
               " G o ahead, don't spare your strength."
               The result was astonishing: there was a crack, the
            ruler broke, but the newspaper remained on the table,
            still covering the other piece of the ruler.
Figure 63

                Alex asked archly, "The newspaper appears to be
            heavier than you've been thinking?"
                Bewildered, I shifted the eyes from the fragment of
            the ruler to the newspaper.
               "Is it an experiment? Electric?"
               "It's an experiment but not electric one. The electric
            ones will follow. I just wanted to show you that
            a newspaper can actually be a device to do physical
            experiments with."
               "But why didn't it let the ruler go? Look, I can easily
            lift it from the table."
               That is the kernel of the experiment. Air presses
            down on the ruler with a powerful force: a good solid
            kilogramme on each centimetre of the newspaper.
            When you strike the protruding end of the ruler, its
            other end pushes up against the newspaper from below,
            and so the newspaper should rise. If it's done slowly
            some air gets under the rising paper and compensates
            for the pressure from above. But your stroke was so
            fast that air had no time to get under the paper. Thus
            the edges of the paper were still sticking to the table
            when its middle was already being forced upwards.
            Therefore, you had to lift not only the paper but also
86-87   A Sheet of   Newspaper

        the paper with the air pressing down on it. In a nut-
        shell, you had to lift with the ruler as many
        kilogrammes as there were square centimetres in the
        newspaper. If it were an area of only 16 square
        centimetres (a square with a side of 4 centimetres), then
        the air pressure would be 16 kilogrammes. But the area
        you lifted was notably larger, and accordingly, you had
        to lift a substantial weight, perhaps something near 50
        kilogrammes. The ruler couldn't bare this load and
        broke. Now do you believe that a newspaper can be
        used for physical experiments? After dark, we'll make
        the experiments."

        Sparks from Fingers• Obedient   Sticks
         Electricity in Mountains

        My brother took a clothes-brush in one hand and held
        the newspaper against the warm stove with the other.
        He then began to rub the newspaper with the brush
        like a decorator smoothing wall-paper on the wall for
        the paper stick perfectly.
           "Look!" Alex said and took both hands away
        from the paper.
           I had expected that the paper would slide down onto
        the floor. This, however, didn't happen: strange as it
        was, the paper stuck to the smooth tiles as if glued.
           I asked, "How does it keep on? It's not smeared with
           "The paper is held by electricity. It's now electrified
        and attracted to the stove."
           "Why didn't you tell me that the newspaper in the
        bag was electrified?"
           "It wasn't. I did it right now, before your eyes, by
        rubbing it with the brush. The friction electrified it."
           "So, it's a real electric experiment?"
           "Yes, but we're just beginning... Turn off the lights
           In the dark the black figure of my brother and the
        greyish spot of the stove looked blurred.
           "Now watch my hand."
           I guessed, rather than saw what he did. He took the
        paper down from the stove and, holding it with one
        hand, moved his spread fingers of the other hand to it.
           And t h e n - I could hardly believe my eyes-sparks
        flew out from his fingers, bluish-white sparks!
           "The sparks were electricity. Want to try for your-
            86-87   A Sheet of   Newspaper

               I promptly hid my hands behind the back. N o t for
Figure 64   the world!
               My brother again applied the paper to the stove,
            brushed it and again produced an avalanche of long
            sparks from his fingers. I managed to notice that he
            didn't touch the newspaper at all, but held his fingers
            about 10 centimetres away from it.
               "Don't be scared, just try, it doesn't hurt. Give me
            your hand," he took hold of my hand and led me to
            the stove. "Spread your fingers!.. Well! Does it hurt?"
               In a twinkling of an eye a bundle of bluish sparks
            shot out from my fingers. In their light I saw that my
            brother had only partially detached the newspaper from
            the stove, the lower part being still "glued". Simultane-
            ously with the sparks I felt a slight prick but the pain
            was trifling. Indeed, nothing to be scared of.
               "Again!" I asked.
               Alex applied the newspaper to the stove and began
            to rub, this time only with his palms.
               "What are you doing? Have you forgotten the
               "It's all the same. Now, are you ready?"
               "Nothing doing! You've rubbed it with bare hands
            without using the brush."
               "It's possible without the brush too if only your
            hands are dry. You just have to rub."
               And it was true, this time also sparks rained from my
               After I had the sparks to my heart's content my
            brother proclaimed: "That'll do. Now I am going to
            show you a flow of electricity, just like the one
            Columbus and Magellan saw at the tops of the masts
            of their ships... Pass me the scissors."
               In the dark Alex brought the points of the spread
Figure 65   scissors near to the newspaper, which was half-
            separated from the stove. I expected to see sparks but
            saw something new, the points of the scissors were
            crowned by glowing bundles of short bluish and
            reddish threads although the scissors were still far away
            from the paper. This was accompanied by faint
            prolonged hissing.
               "Sailors often see the same sort of fire brushes, only
            far larger ones, at the mastheads and yardarms. They
            called them St. Elm's fires."
               "Where do they come from?"
               "You mean who holds an electrified newspaper above
            the masts? True, there's no newspaper there, but a low
            86-87   A Sheet of   Newspaper

            electrified cloud. It's a substitute for the newspaper.
            You shouldn't think, however, that this sort of electric
            glow on pointed structures only occurs at sea, it's also
            observed on land, especially in the mountains. So,
            Julius Caesar wrote that on a night in a cloudy weather
            the spear heads of his legionnaires glowed this way.
            Sailors and soldiers are not afraid of these electric
            l i g h t s - o n the contrary they view them as a good omen.
            Of course, without any reasonable grounds. In the
            mountains electric glows even occur on people at times,
            on their hair, caps, e a r s - t h a t is, on all the protruding
Figure 66   parts. In the process, they often hear a buzz, like the
            one produced by our scissors."
                 "Does this fire burn strongly?"
                 "Not at all. After all, this isn't a fire, but a glow, just
            a cold glow. So cold and harmless that it cannot even
            ignite a match. Look: instead of the scissors I use
            a match. And you see: the head is surrounded by the
            electric glow, but it doesn't go off."
                 "But I think it is burning because flames are coming
             out straight from the head."
                 "Turn on the light and inspect the match."
                 I made sure that the match not only hadn't charred
            but it hadn't even blacken. It was thus indeed sur-
             rounded by a cold light, and not fire.
                 "Leave the light on. We'll carry out the next
            experiment in the light."
                 Alex shifted a chair to the middle of the room and
             put a stick across its back.
                 After several tries he managed to balance the stick at
            one point.
                "I didn't know that a stick could be supported in this
            way," I said, "it's so long."
                "It works for exactly that reason. A short one
            wouldn't. A pencil, for example."
                I agreed, "A pencil, no means."
                "Now, to business. Can you make the stick turn
            towards you without touching it?"
                I thought about it.
                "If we loop a rope onto its end..." I began.
                " N o ropes, it must touch nothing. Can you?"
                "Aha, insight!"
                I put my face close to the stick and began sucking air
            into my mouth to attract the stick to me. It didn't stir,
                "Any progress?"
                "None. It's impossible!"
            86-87   A Sheet of   Newspaper

               "Impossible? Let's see."
               He took the newspaper, down from the stove, where
            that had been sticking to the tiles, and began slowly to
            move it sidewards towards the stick. At about half
            a metre away the stick "felt" the attraction of the
            electrified newspaper and obediently turned in its
            direction. By moving the newspaper Alex made the
Figure 67
            stick follow it rotating at the back of the chair, first in
            one direction, then in the other.
               "The electrified newspaper, you see, attracts the stick
            so strongly that it follows and will follow the paper
            until all the electricity has flowed from the newspaper
            into the air."
               "It's a pity that these experiments cannot be
            performed in the s u m m e r - t h e stove will be cold."
               "The stove is only necessary to dry up the paper
            since these experiments are only a success with an
            absolutely dry newspaper. You may have noticed that
            newspapers absorb moisture from the air and therefore
            are always somewhat damp, that's why it has to be
            dried. You shouldn't think, however, that in summer
            our experiments are impossible. They can be done but
            not so well as in winter when the air in a heated-up
            room drier than in summer - that's the reason. Dryness
            is crucial for these experiments.              In     summer,
            a newspaper can be dried with a kitchen stove when it's
            not too hot for the paper not to ignite on it. After the
            paper has been dried adequately, it's brought onto
            a dry table and rubbed hard with a brush. The paper
            electrifies, but not as with a tile stove... Well, let's call it
            a day. Tomorrow we'll do some new experiments."
               "Also electric?"
               "Yes, and with the same electric machine, our
            newspaper. Meanwhile I'll give you an interesting
            account of Elm's fires by the famous French naturalist
            Saussure. In 1867 he with several companions climbed
            the Sarley Mountain, which is more than 3 kilometres
            high. And here's what they experienced there.
               Alex took down the book The Atmosphere                    by
            Flammarion from the bookcase, thumbed through it
            and gave me the following passage to read:
               "The climbers leaned their alpenstocks against a cliff
            and were preparing for their dinner when Saussure felt
            a pain in his shoulders and back as if a needle were
            being driven slowly into his body. Thinking some pins
            had got into my canvas cape,' recounted Saussure, 'I
            threw it off but there was no relief, on the contrary, the
86-87   A Sheet of   Newspaper

        pain became more accute and embraced the whole of
        my back from shoulder to shoulder. It was as if a wasp
        was walking all over my skin stinging it everywhere.
        I hastily threw off another coat but I could find
        nothing that could hurt so badly. The pain continued
        and came to feel like a burn. It seemed to me that my
        woollen sweater had caught fire and I was about to
        undress when my attention was attracted by a noise,
        a sort of hum. It came from the alpenstocks we had
        leaned against the cliff and resembled the rumbling of
        heated water about to boil. The noise continued for five
        minutes or so.
           'I then understood that the painful sensation was
        caused by an electrical flux released by the mountain.
        In the broad daylight I didn't see any glow on the
        alpenstocks. They produced the same sharp noise
        whether they were held vertically with the tip pointed
        up and down, or horizontally. No sound came from the
           'In several minutes I felt that the hair on my head
        and beard were rising as if a dry razor was being
        passed over a stiff beard. My young companion cried
        out that the hair of his moustache was rising and the
        tops of his ears were giving off strong currents. Having
        raised my hands, I felt currents emanating from my
        fingers. In short, electricity was being liberated from
        sticks, clothing, ears, hair, in fact everything that was
           'We left the summit hastily and descended about
        a hundred metres. As we were climbing down, our
        alpenstocks were producing ever lower noise and
        finally, the sound became so soft that we could only
        hear it by bringing them close to our ears.'"
           In the same book I read about other cases of Elm's
           "The liberation of electricity by protruding rocks is
        often observed when the sky is covered by low clouds
        gliding just over summits.
           "On June 10, 1863, Watson and several tourists
        climbed up the Jungfrau pass in Switzerland. It was
        a fine morning but the travellers got into a strong hail
        storm in the pass. A terrible clap of thunder came and
        soon Watson heard a hissing sound coming from his
        stick that resembled the sound of a kettle about to boil.
        The people stopped and found that their sticks and
        axes produced the same sound, and didn't stop making
        the sound even when stuck with one end into the
            A Sheet of   Newspaper

            ground. One of the guides took his hat off and cried
            that his head was burning. His hair stood on end as if
            electrified and everybody had tickling feelings on the
            face and other parts of the body. Watson's hair
            straightened out completely. The stirring of fingers in
            the air produced electric hiss from their tips."

            Dance of Paper Buffoons • Snakes   •
            Hair on End
            Alex kept his word. Next day after dark he resumed the
            experiments. First of all he "glued" the newspaper to
            the stove. Then he asked me to get him paper denser
            than newspaper, writing paper for example, out of
            which he cut out some funny figures, small dolls in
            various stances.
               "These paper buffoons will now dance. Fetch me
            a few pins."
               Soon each buffoon's foot was pinned up.
Figure 68
               "This is for the buffoons not to be scattered and blown
            away by the newspaper," Alex proclaimed arranging the
            figures on a tray, "The performance starts!"
               He "unglued" the newspaper from the stove and,
            holding it horizontally with both hands, brought it
            down to the tray with the figures.
               Alex commanded, "Stand up!"
               And just imagine: the figures obeyed. They stood up
            and stayed that way until he removed the newspaper
            when they lay down again. But he didn't allow them to
            rest long: by alternately moving the newspaper to and
            from them he made the buffoons stand up and lie down
Figure 69      "If I hadn't burdened them with pins, they would
            have jumped up and stuck to the newspaper." My
            brother took the pins out of some of the figures, "You
            see, they've stuck to the newspaper and won't separate
            from it. This is electric attraction. And now we'll
            experiment with electric repulsion, too. Where are the
               I passed them to him and, having "glued" the
            newspaper to the stove, Alex began to cut a long, thin
            strip from its lower edge almost up to the upper one.
            Similarly, he made a second, third and other strip.
            When he got to about the sixth strip, he cut all the way
            to the edge. He had thus produced a sort of paper
            beard that didn't slide down from the stove, as I had
            expected, but stayed put. Holding the upper part with
            one hand, Alex brushed along the strips several times
94-95       86-87   A Sheet of   Newspaper

            and then took the "beard" down from the stove holding
            it at the top.
               Instead of freely dangling the strips spread out into
            a bell shape noticeably repelling one another.
Figure 70      My brother explained, "They repel one another
            because they are charged up identically. However they
            are attracted to the things that are not charged. Poke
            your hand into the bell from below and the strips will
            be attracted to it."
               I sat down and put my hand into the space between
            the strips. That is, I wanted to poke my hand there but
            couldn't because the paper strips wound themselves
            round my hand like snakes.
               "Aren't you frightened by these snakes?" asked Alex.
               "No, they are only paper."
               "But I'm scared. Look, how scared."
               Alex raised the newspaper above his head and I saw
            his long hair literally stand on end.
               "Is it another experiment?"
               "The same experiment we've just done, only another
            form of it. The newspaper electrified my hair and it is
            now attracted to the newspaper while each piece of hair
            repels the others like the strips of our paper beard.
            Look in the mirror and I'll show you your own hair
            standing on end in the same fashion,"
               "Does it hurt?"
               "Not a bit."
               Really, I felt not the slightest pain, not so much as

Figure 71

            tickling, although the mirror clearly showed to me that
            my hair under the newspaper stood on end.
               In addition, we repeated yesterday's experiments and
            then my brother discontinued the "session", as he called
            it, promising to do some new experiments tomorrow.
            86-87   A Sheet of   Newspaper

            Miniature Lightning 0 Experiment   with a Water Stream   •
            Herculean   Breath

            On the next night my brother Alex made some unu-
            sual preparations.
               He took three glasses, warmed them at the stove,
            put them on the table and covered with a tray that he
            had also preheated.
               "What is it going to be?" I inquired. "Shouldn't the
            glasses be placed on the tray and not vice versa?"
               "Just wait, take your time. It's going to be an
            experiment with miniature lightning."
               Alex used the "electric machine" a g a i n - t h a t is, he
            simply rubbed the newspaper on the stove. He then
            folded the newspaper in two and resumed the rubbing.
            Next he "unglued" it from the stove tiles and swiftly
            put on the tray.
               "Feel the tray... Is it cold?"
               Suspecting nothing, I light-heartedly stretched out
            my hand... and promptly jerked it back: something had
            cracked and pricked my finger painfully.
               Alex laughed, "How did you like it? You were struck
            by a lightning. Heard the crack? That was miniature
               "I felt a strong prick, but I didn't see any lightning."
               "You will when we repeat it in the dark."
              "But I won't touch that tray any more," I proclaimed
              "That's not necessary, we can produce sparks, say,
            with a key or a tea-spoon. You'll feel nothing, but the
            sparks will be as long as they were earlier. The first
            sparks I'll extract myself while your eyes adapt to the
Figure 72
               He turned off the light.
              "Silence now. Keep your eyes open!" a voice said in
            the dark.
              Crack! and a bright bluish-white spark about 20
            centimetres long darted between the edge of the tray
            and the key.
               "See the lightning? Hear the thunder?"
              "But they were at the same time. A real thunder
            always comes after you see the lightning."
              "True. We always hear thunder later. Still, they occur
            simultaneously, rather like the crack and spark in our
              "Why then is the thunder later?"
              "You see, lightning is light, and it travels so fast as to
96-97       86-87   A Sheet of   Newspaper

            cover terrestrial distances in almost no time. Thunder is
            an explosion, i. e. sound, but sound travels in air not so
            fast and markedly lags behind the light, thus coming to
            us later. That's why we see a lightning flash before we
            hear the accompanying thunder."
               Alex passed me the key, removed the newspaper
            a n d - m y eyes had now adapted - suggested extract
            a "lightning" from the tray.
               "Without the newspaper, will there be any spark?"
               "Just try."
               I had hardly put the key near the tray edge when
            I saw a spark, long and bright.
               My brother again put the newspaper on the tray and
            again I extracted a spark, though it was weaker this
            time. He did so dozens of times (without rubbing the
            newspaper again on the stove), and each time I made
            a spark, which was getting ever weaker.
               "The sparks would continue for a longer time if
            I held the newspaper silk strings or ribbons rather than
            with my bare hands. When you study physics you'll
            understand what occurred. Meanwhile it only remains
            for you to look with your eyes not head. Now one
            more experiment, with a water stream. We'll make it in
            the kitchen at the water tap. Let the newspaper stay on
            the stove."
               We make a thin stream of water from the tap so that
            it hits the basin bottom loudly.
               "Now, without touching the stream, I'll make it fall
            somewhere else. Which way do you want it to be
            deflected, left, right, or forward?"
               I replied at random, "To the left."
Figure 73      "All right. Don't touch the tap while I fetch the
               Alex was back with the newspaper, trying to hold it
            with his arms outstretched so as not to lose too much
            electricity. He brought the newspaper close to the
            stream from the left and I clearly saw it bend to the
            left. Having transferred the newspaper on the other
            side, he made the stream deflect to the right. Finally, he
            drew it forward so much that the water poured over
            the basin edge.
               "You see how strongly the attractive force of
            electricity manifests itself. By the way, this experiment
            can also be easily performed without a stove or oven.
            If, instead of the charged newspaper you take
            a conventional plastic comb like this one," my brother
            produced a comb and passed it through his thick hair.
            A Sheet of   Newspaper

            "I have charged it up this way."
                "But your hair is not electrical."
                "No, it's just like yours or anybody else's. But if you
            rub plastic on your hair, it gets charged in the same
            way the newspaper does by the brush. Look."
                When the comb was brought to the water stream, it
            made it deflect noticeably.
                "The comb is unsuitable for other our experiments
            since it accumulates too little electricity. It gets far less
            than the 'electric machine' that can be made from
            a simple newspaper. I'd like to make one more
            experiment with the newspaper, the last one. This time
            it's not an electric experiment, but again one with air
            pressure, rather like the experiment with the ruler."
                We returned to the sitting room and Alex began to
            cut and glue a long bag out of a newspaper.
                "While it dries, get several books, large and heavy."
                On the bookshelf I found three massive volumes of
Figure 74
            some medical atlas and placed them on the table.
                My brother asked, "Can you inflate this bag with
            your mouth?"
                "Of course."
                "A simple business, isn't it? But what if I put
            a couple of these volumes on the bag?"
                "Oh, then the bag won't inflate no matter how hard
            you try."
                Alex silently put the bag at the edge of the table,
            covered it with one of the volumes and stood another
            one upright on it.
                "Just watch. I'll inflate the bag."
                "Perhaps you want to blow those books away?"
            I asked laughing.
                Alex started to blow into the bag. Just imagine: as
            the bag swelled the lower book sloped up and
            overturned the top one. But the two books weighed
            about five kilograms!
                Without allowing me to recover from my surprise,
Figure 75
            Alex prepared to repeat the trick. This time he loaded
            the bag with all three tomes. He b l e w - a Herculian
            breath!-and the three tomes overturned.
                The amazing thing is that this experiment had
            nothing miraculous in it. When I dared to repeat it for
            myself, I managed to overturn the books as easily as
            Alex did. You need to have neither an elephant's lungs
            nor the muscles of an athlet-everything comes about
            on its own accord, nearly without effort.
86-87   A Sheet of   Newspaper

           My brother later explained the reason to me. When
        we inflated the paper bag, we forced some air into it
        that is more compressed than the air around us,
        otherwise the bag wouldn't expand. The air outside
        presses down with about 1,000 grammes on each square
        centimetre. If you express the area under the books in
        square centimetres, you can readily work out that even
        if the excess pressure in the bag is only a tenth of that
        outside of it, i.e. a hundred grammes per square
        centimetre, then the total force from the air pressure
        inside the part of the bag under the books may be as
        high as 10 kilogrammes. Clearly this force is sufficient
        to overturn the books.
           Thus ended our physics tests with the newspaper.
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            How to Weigh Accurately      with
            An Inaccurate Balance
            Which is the more important possession, a precise pair
            of weighing scales or a precise set of weights? Many
            people believe that the scales are more important, but
            in f a c t - t h e weights, since it's impossible to weigh
            anything accurately with inaccurate weights. If the set
            of weights is true, then you can still weigh quite
            accurately with inaccurate scales.
              For example, suppose you have beam scales with
            pans. Place a weight that is heavier than your object on
            one pan. Then on the other pan put as many weights as
            will be required to make the beam balance. Next put
            your object onto the pan with the weights and, of
            course, this pan will sink. In order to balance the beam
            again you will need to remove some of the weights and
            the weights removed will show the correct weight of
            your object. It should be clear why. your object now
            pulls down its pan with the same force with which the
            weights you took off did before. Hence your object and
            the total of the weights you took off weigh the same.
              This excellent way of weighing accurately using
            inaccurate scales was discovered by the great Russian
            chemist Dmitri Mendeleyev.

            On the Platform of a Weighing       Machine
            A man stands on the platform of a weighing machine
            and suddenly he squats down. Which way will the
            platform move, up or down?
              The platform will move upwards. Why? Because as
            he is squatting the muscles pulling the man's body
Figure 76   down also pull the legs up, thus reducing the force with
            which the body presses on the platform with the result
            that it goes up.

            Weight on Pulley
            Suppose a man is able to lift a mass of 100
            kilogrammes from the floor. Wanting to lift more he
            passed a rope tied tc the load through a pulley fixed in
            the ceiling (Fig. 76). What load will he be able to lift
            using this rig?
               Such a pulley could help him lift no more than what
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          he could do with his own hands, perhaps even less. If
          he pulled the rope passed through a fixed pulley, he
          could not lift a mass exceeding his own. If his mass is
          less than 100 kilogrammes, he would be unable to
          handle a 100-kilogramme load with the pulley.

          Two Harrows
          People often confuse weight and pressure. However,
          they are by no means the same. An object may have
          a marked weight but still exert a negligible pressure on
          its support. By contrast something else may have
          a small weight but exert a large pressure on its support.
             The following example will clarify the difference
          between weight and pressure and at the same time give
          you an idea of how to work out the pressure a body
          exerts on its support.
             Let two harrows of the same type work in field, one
          with 20 teeth, the other with 60, the first one weighing
          60 kilogrammes, the second 120 kilogrammes.
             Which one penetrates more deeply into the soil?
             It's easy to figure out that the greater the force acting
          on a harrow's teeth, the deeper they penetrate the soil.
          With the first harrow the total load of 60 kilogrammes
          is evenly distributed among the 20 teeth, hence
          3 kilogrammes per tooth. With the second harrow,
          120/60, i.e. 2 kilogrammes per tooth. Consequently,
          though in general the second harrow is heavier, its
          teeth penetrate less deeply than the first harrow's. The
          pressure per tooth with the first harrow is larger than
          with the second.
          Pickled Cabbage
          Consider another simple calculation of the pressure.
             Two barrels of pickled cabbage are each covered
          with a wooden disk held down by stones. One disk is
          24 centimetres across and the stones on it weigh 10
          kilogrammes while the other is 32 centimetres across
          and its stones weigh 16 kilogrammes.
             In which is the pressure larger?
             Clearly, the pressure will be higher in the barrel
          where the load per square centimetre is larger. In the
          first case the 10 kilogrammes are distributed over an
          area * of 3.14 x 12 x 12 = 452 square centimetres.

              * The area of a circle is about 3.14 times the circle's radius (half
          the diameter) times the circle's radius.
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Hence the pressure is 10,000/452, i.e. about 22 grammes
per square centimetre. In the second barrel the pressure
will be 16, 000/804, i.e. less than 20 grammes per square
centimetre. The pickled cabbage is thus more
compressed in the first barrel.

Awl and Chisel
Why does an awl penetrates deeper than a chisel does if
both are acted upon by an equal force?
   The point is that when thrusting the awl all the force
is concentrated at an extremely small area at its point.
With the chisel the force is distributed over a much
larger surface. For instance, let the awl's surface area at
the point be 1 square millimetre and the chisel's be
1 square centimetre. If the force on each tool is one
kilogramme, then the material under the chisel blade is
subjected to a pressure of 1 kilogramme per square
centimetre, and under the awl 1/0.01 = 100 or 100
kilogrammes per square centimetre (since 1 square
millimetre = 0.01 square centimetre). The pressure of
the awl is one hundred times larger than of the chisel.
Now it is clear why the awl penetrates deeper than the
   You'll now understand that when you are pressing
with your finger on a needle when you are sewing you
produce a very great pressure, not smaller than the
steam pressure in a boiler. This is also the principle
behind the cutting action of the razor. The slight force
of hand creates a pressure of hundreds of kilogrammes
per square centimetre on the thin edge of the razor that
can cut through hair.

Horse and Tractor
A heavy crawler tractor is well supported by loose
ground into which the legs of horses and people are
mired. This is inconceivable to many people since the
tractor is far heavier than the horse and very much
heavier than man. Why then do the horse's legs are
mired in loose ground, and the tractor doesn't?
   To grasp this, you'll have to remember once again
the difference between weight and pressure.
   An object does not penetrate deeper because it is
heavier but because it exerts a higher pressure (or force
per square centimetre) on its support. The enormous
weight of a crawler tractor is distributed over the larger
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            surface area of its tracks. Therefore, each square
            centimetre of the tractor's support carries a load as low
            as several grammes. On the other hand, the horse's
            weight is distributed over the small area under its
            hooves, thus giving more than 1,000 grammes per
            square centimetre or ten times more than the tractor.
            N o wonder then that a horse's feet sink more deeply
            into mud than does a heavy crawler tractor. Some of
            you may have seen that to ride over marshes and bogs
            horses are shod with wide "shoes", which increase the
            supporting area of horses' hooves with the result that
            they are mired much less.

            Crawling Over Ice
            If ice on a river or lake is insecure, experienced people
            crawl rather, than walk over it. Why?
               When a man lies down, his weight, of course, doesn't
            change, but the supporting area increases, each square
            centimetre of it thus carries less load. In other words,
            the man's pressure on his support is reduced.
               It's now clear why it's safer to move over thin ice by
            crawling - this decreases the pressure on the ice. Some
            people also use a wide board and lie on it as they move
            about thin ice.
               What load can ice support without breaking? The
            answer is dependent on the thickness of the ice. Ice
            4 cm thick can support a walking man.
               It is of interest to know the thickness of ice required
            for a skating rink on a river or lake. For this purpose
            10-12 centimetres would be sufficient.
Figure 77

            Where Will the String         Break?
            You'll need an arrangement shown in Fig. 77. Put
            a stick on top of the open doors, tie a string to the
            stick and tie a heavy book in the middle. If now you
            pull a ruler tied to the bottom of the string, where will
            the string break: above or below the book?
               The string can break both above and below,
            according as you pull. It's up to you, you can break it
            either way. If you pull carefully, the upper part of the
            string will break, but if you jerk it, the lower part
               Why does this happen? Careful pulling breaks the
            upper part of the string because the string is being
            pulled down both by the force of your hand and by the
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            weight of the book, whilst the lower part of the string is
            only acted upon by the force of your hand. Whereas
            during the short instant of the jerk the book doesn't
            acquire very much motion and therefore the upper part
            of the string doesn't stretch. The entire force is thus
            "consumed" by the lower part, which breaks even if it's
            thicker than the upper part.

            Torn Strip
Figure 78   A strip of paper that is about 30 cm long and one
            centimetre wide can be material for a funny trick.
            Partly cut or tear the strip in two places (Fig. 78) and
            ask your friend what will happen to it if it's pulled by
            the ends in the opposite directions.
               He will answer that it'll break in the places where it's
            been torn.
               "Into how many parts?" you ask then.
               Generally the answer is: "Into three parts, of course."
            If you receive this answer, ask your friend to test his
            hunch by an experiment.
               Much to his surprise he will see that he was
            mistaken, for the strip will only separate into two parts.
               You can repeat the experiment many times taking
            strips of various length and making little tears of
            various depth and you'll never get more than two
            pieces. The strip breaks where it's weaker which goes to
            prove the proverb: "The chain is only as strong as its
            weakest link". The reason is that of the two tears or
            cuts, however hard you strive to make them identical,
            one is bound to be deeper than the other. Even if it's
            imperceptible to your eyes, one will still be deeper. The
            weakest place of the strip will be first to begin to break.
            And once begun the breaking would continue to the
            end because the strip would become ever weaker at this
               You might perhaps be very pleased to know that in
            making this trifling experiment you've visited a serious
            branch of science of importance for engineering that is
            called "strength of materials".

            A Strong Match      Box
            What will happen to an empty match box if you strike
            it with all your might?
               I'm sure that nine out of ten readers will say that the
            box won't survive such handling. The tenth - the person
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            who has actually performed the experiment or heard
Figure 79   about it from somebody-will maintain that the box
            will survive.
               The experiment should be staged as follows. Put the
            parts of an empty box one on top of the other, as
            shown in Fig. 79. Strike this assembly sharply with
            your fist. What will occur will surprise you: both parts
            will fly apart but, having collected them, you'll find that
            each one is intact. The box behaves like a spring and
            this saves it because it bends but doesn't break.

            Bringing Something Closer by Blowing
            Place an empty match on a table and ask somebody to
            move it away by blowing. Clearly this is no problem.
            Then ask him or her to do the opposite, i.e. make the
            box approach without leaning forward to blow the box
            from behind.
              There'll hardly be many who'll twig. Some will try to
            move the box nearer by sucking in air, but that won't
            work of course. The answer, however, is very simple.
              What is it?
               Ask somebody to put the hand vertically behind the
            box. Begin to blow and the air that reflects from the
            hand will strike the box and shift it towards you
            (Fig. 80).
Figure 80

              The experiment is, so to speak, "failsafe". You'll only
            have to make sure it's on a sufficiently smooth table
            (even unpolished) which is not of course covered with
            a table-cloth.

            Grandfather's   Clock
            Suppose a grandfather's clock that uses weights to wind
            it up is fast or slow. What should be done with the
            pendulum to correct it?
               The shorter a pendulum the quicker it swings. You
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            can easily prove this by an experiment with a weight
            suspended from a piece of rope. This suggests the
            solution of our problem: when the clock is slow you
            shorten the pendulum a little by lifting a ring on the
            pendulum rod, so making the pendulum swing faster;
            and when the clock is fast you lengthen the pendulum.

            How Will Rod Settle Down?
Figure 81   Two balls of equal mass are fixed to the ends of a rod
            (Fig. 81). Right in the middle of the rod a hole is
            drilled through which a spoke is passed. If the rod is
            spun about the spoke, it'll rotate several times and
            settle down.
               Could you predict in what position the rod will come
            to rest?
               Those who think that the rod will invariably
            settle down in a horizontal position are mistaken.
            It can remain balanced in any position (see
            Fig. 81)-horizontal, vertical, or at an angle-since it's
            supported at the centre of mass. Any body supported
            or suspended at the centre of mass be in equilibrium at
            any position. Therefore, it's impossible to predict the
            final position of the rod.

            Jumping in Railway      Carriage
            Imagine you are travelling in a train at a speed of 36
            kilometres an hour and you jump up. Supposing that
            you manage to spend a whole second in the air (a
            brave assumption because you'll need to jump up more
            than a metre), where will you land, at the same place
            from where you started or somewhere else? If
            somewhere else, where then-closer to the beginning or
            end of the train?
               You'll land at the same place. You shouldn't think
            that while you've been in the air the floor (together
            with the carriage) has shifted forward. To be sure, the
            carriage was tearing along but you also were travelling
            in the same direction and at the same speed carried by
            inertia. All the time you were directly above the place
            from which you jumped up.

            Aboard a Ship
            Two people are playing ball on the deck of a steaming
            vessel (Fig. 82). One stands nearer the aft and the other
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Figure 82

            nearer the bows. Which one can throw the ball easier
            to his partner?
               If the ship is travelling with a steady speed and in
            a straight line neither has any advantage, just as if they
            were on a stationary ship. You should not suppose that
            the man standing nearer the bows recedes from the ball
            after it's been thrown or that the other man moves to
            meet it. By inertia the ball has the ship's speed which is
            equally possessed by both partners and the ball. There-
            fore, the motion of the ship (uniform and rectilinear)
            gives neither player an advantage.

            A balloon is being carried away due north. In which
            direction will flags on its car fly?
               The balloon carried by an air flow is at rest with
            respect to the surrounding air, therefore the flags won't
            be blown by the wind, but will dangle limply like they
            do in still weather.

            On a Balloon
            A balloon floats motionlessly in the air. A man gets out
            of the car and begins to climb up the cable. Which way
            will the balloon move in the process, upwards or
              The balloon will shift downwards, since the man
            pushes the cable (and the balloon) in the opposite di-
            rection as he is climbing. The situation is similar to what
            happens if someone walks forward over the bottom
            of a small rowing boat: the boat shifts backwards.
            Walking and Running
            What is the difference between running and walking?
             Before answering remember that running can be
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            slower than walking and that you can even run on the
               The difference between running and walking is not
            the speed. When we walk our body is in contact with
            the ground all the time at some point in our feet. When
            we run, on the other hand, there are moments when the
            body is completely separated from the ground and does
            not touch it at any point.

            A Self-Balancing    Stick
            Put a smooth stick on the index fingers of both of your
            hands, as shown in Fig. 83. Now move your fingers
            together to meet each other half-way. Strangely, in the
            final position the stick doesn't fall off but keeps its
            balance. Make the experiment several times varying the
            initial position of your fingers, the result will invariably
            be the same: the stick will be balanced each time.
            Replace the stick by a ruler, a billiards cue, or a broom,
            and you'll notice the same behaviour.

Figure 83

               What is the secret?
               The following is clear: if the stick is balanced on
            your fingers brought together, this suggests that your
            fingers have closed up under the centre of mass (a body
            is in equilibrium if the centre of mass is over an area
            confined by the support's boundaries).
               When your fingers are spread apart, the larger load is
            on the finger that is closer to the stick's centre of mass.
            Friction increases as the load grows and the finger
            closer to the centre of mass is subject to larger friction
            than the other one. Therefore, the finger that is closer
            to the centre of mass doesn't slide under the stick and
            at all times the only finger that moves is the one farther
            away from this point. Once the moving finger is closer
            to the centre of mass than the other, the fingers change
            their roles, the change taking place several times until
            the fingers come together. Since only one finger is
            moving at each instant of time, namely the one that is
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            farther away from the centre of mass, it's only natural
            that eventually both fingers end up under the centre of
            mass of the stick.
               Before we leave this experiment, we'll repeat it with
            a broom (the top of Fig. 84) and ask ourselves what
            would happen if we cut the broom at where it's
            supported by the fingers and place the parts on
            different pans, which pan would sink?
Figure 84

              It would seem that if both parts of the broom
            balance each other on your fingers these should also do
            so on the pans of the scales. Actually, the part with the
            brush will outweigh. The clue is not difficult to find, if
            we take into account that when the broom was
            balanced on your fingers the gravity forces of both
            parts were applied to unequal arms of the lever. On the
            pans of the scales, by contrast, the same forces are
            applied to the ends of an equal-arm lever.

            Rowing in the River
            A rowing boat and wooden chip alongside it are
            floating in a river. What is it easier for the rower: to
            get ahead of the chip by 10 metres or to lag behind it
            by 10 metres?
               Even those practising water sports often give the
            wrong answer. It's more difficult, they argue, to row
            upstream than downstream, accordingly, to pass the
            chip is easier than to lag behind it.
               No doubt, to reach a point rowing upstream is more
            difficult than rowing downstream. But if the point you
            are going to reach is floating alongside, just like our
            chip, the situation is quite different.
               One should take into account that the boat carried
            by the current is at rest with respect to the water. The
            situation is the same as what it would be on the still
            water of a lake.
               Thus, in both cases the rower needs exactly the same
            effort whether he wishes to pass or lag behind the boat.
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Circles on Water
A stone thrown into still water produces concentric
   What form will the waves have if the stone is thrown
into the flowing water of a river?
   If you fail, from the very beginning, to follow the
right track, you'll be easily lost in the argument and
come to the conclusion that in the flowing water the
waves should assume the form of an ellipse or an
oblong somewhat wider in the upstream direction. But
if we attentively observe the waves produced by a stone
thrown into the river, we'll find no deviation from the
circular shape, however fast the stream is.
   There is nothing extraordinary in that. Simple
reasoning will lead to the conclusion that the waves
should be circular both in still and in flowing water.
Let's treat the motion of particles of the waves as
a combination of two movements: radial (from the
centre of oscillations) and translational (downstream).
A body participating in several motions eventually
comes to the same point it would come to, if it
performed all the component motions in succession.
   We'll therefore assume that the stone is thrown into
still water. In that case, the waves will clearly be
   Now suppose that the water is m o v i n g - n o matter
with what velocity, uniformly or n o t - t h e motion has
only to be translational. What will happen to the
circular waves? They'll undergo a translation without
any distortion, i.e. will remain circular.

Deflection of Candle Flame
If you carry a candle about a room you will have
noticed that initially the flame deflects backwards.
Which way will it deflect if the candle is carried about
in a closed casing? Which way will the flame in the
casing deflect if it's uniformly rotated, horizontally in
an outstretched hand?
   If you think that in the casing there'll be no
deflection, you are mistaken. Experiment with
a burning match and you'll see that if it's protected by
the hand as it's moved, the flame will deflect, but
forwards-quite unexpectedly!-not backwards. This is
because the flame is thinner than the surrounding air.
A force imparts to a body with a small mass a larger
velocity than to a body with a larger mass. Therefore,
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          the flame moving faster than the air in the casing
          deflects forwards.
             The same reason (the smaller density of the flame
          than that of the surrounding air) also accounts for the
          unexpected behaviour of the flame when we move the
          casing in a circle, the flame deflects inwards, not
          outwards as might be expected. This would be clear if
          we remember how mercury and water are arranged in
          a ball rotated in a centrifuge. The mercury tries to be
          farther away from the rotation axis than the water. The
          latter, as it were, floats up in the mercury, if we
          consider the "bottom" to be the direction away from
          the rotational axis (i.e. the one in which bodies are
          displaced by the centrifugal effect). In our circular
          rotation, the lighter-than-air flame "floats up" within
          the casing, i.e. in the direction to the rotation axis.

          A Sagging Rope
          With what force must one pull at a rope for the latter
          not to sag?
             However taut the rope is, it is bound to sag. Gravity
          that causes the sagging acts normally, whereas the
          stretching force on the rope has no vertical component.
          Two such forces can never balance each other out, i.e.
          their resultant force cannot be zero. And this resultant
          force is responsible for the sagging.
             No force, however strong, can stretch a rope strictly
          horizontally (except when the rope is upright). The
          sagging is unavoidable, you can reduce it to a desired
          degree but cannot make it zero. Consequently, any
          nonvertically stretched rope or driving belt will sag.
             For the same reason it is impossible, by the way, to
          stretch a hammock so that its ropes are horizontal. The
          taut net of a bed sags under the weight of a man. And
          the hammock, whose ropes are not so taut, turns into
          a dangling bag when a man lies on it.

          How to Drop a Bottle?
          In which direction with respect to a moving railway
          carriage should you throw a bottle so that the danger
          that it gets broken when hitting the ground is the least?
            As it is safer to jump forwards from a moving
          carriage, it would appear that the bottle would not hit
          the ground so strongly if you throw it forwards. This is
          not so: objects should be thrown backwards. In that
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            case the velocity imparted to the bottle by throwing it
            will be subtracted from the one due to inertia with the
            result that the bottle will strike the ground with
            a smaller velocity. Throwing the bottle forwards will
            cause the reverse, the velocities would add up and the
            collision would be stronger.
              That it is safer for a man to jump forward is
            accounted for by quite a different reason: he is hurt less
            by jumping this way.

            A piece of cork has got into a bottle with water. The
            cork is small enough to pass freely through the neck,
            but try as you can, shaking or upending the bottle, the
            outpouring water will not for some reason bring the
            cork out. It's only when the bottle is completely empty
            that the cork leaves the bottle with the last bit of the
            water. Why?
                The water doesn't bring the cork out for the simple
            reason that cork is lighter than water and therefore is
            always on its surface. The cork can only come to the
            opening when almost all of the water has come out.
            That's why it is the last to leave the bottle.

            During a spring flood the surface of a river becomes
            convex - higher in the middle than near the banks. If
            loose logs float along such a swollen river, they will
            slide down to the banks leaving the mainstream free
            (the top of Fig. 85). In midsummer, when the water is
            low, the river surface becomes concave - lower in the
            middle than near the banks. In this case logs will accu-
            mulate in the middle (the bottom of Fig. 85).
               What's the reason?
               This is explained by the fact that in the middle water
            flows quicker than near the banks because the friction

Figure 85
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            of the water along the bank slows the current down.
            During a flood, water comes from the upper reaches
            faster along the middle than near the banks because the
            current's speed is faster in the middle. Understandably,
            if more water comes to the middle, then the river
            should swell here. The situation changes in midsummer
            when water subsides. Now, owing to the swifter current
            the water run-off in the middle is higher than near the
            banks with the result that the river becomes concave.

            Liquids...   Press Upwards!
            That liquids exert a pressure downwards, on the
            bottom of a vessel, and sidewards, on its walls, is
            known even to those who have never studied physics.
            But many people don't suspect that liquids press
            upwards as well. A glass tube will help you to make
            sure such a pressure does exist. Cut out a disk of
            a strong cardboard, its size being sufficient to cover the
            hole of the tube. Put it over the hole and dip both into
            some water. For the disk not to drop off when you are
            dipping it, it can be held by a piece of string passed
            through its centre or pressed on with a finger. When
            the tube has sunk to a certain depth, you will notice
            that the disk holds securely on its own without being
            pressed on with the finger or held with the string, being
            supported by the water that presses it up.
               You can even measure the amount of this upward
            pressure. Carefully pour some water into the tube.
            Once the water level approaches the level outside the
            tube, the disk will fall off. The water pressure on the
            disk from below is thus balanced out by the pressure of
            the water column within the tube, its height being equal
            to the depth of the disk in the water. This is a law
            about the pressure of liquids on a submerged body. By
            the way, this also causes the weight "loss" in liquids
            and was formulated as the famous Archimedes

Figure 86
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               If you have several glass tubes of various shapes but
            with the same opening (e.g. as shown in Fig. 86) you
            can also test another law relating to liquids, namely the
            pressure of a liquid on the bottom of a vessel is only
            dependent on the bottom area and the level and is
            independent of the vessel shape. An experiment with
            the various glass tubes is described below. Dip them
            into the water to the same depth (for which purpose
            you'll have to glue paper strips onto them at the same
            height) and you'll notice that the disk will always fall
            off at the same level of the water within the tube (Fig.
            86). In consequence, the pressure due to water columns
            of various shapes is the same if only their base areas
            and heights are the same. Notice that it's the height, not
            the length, that matters because a long inclined column
            exerts exactly the same pressure on the bottom as
            a short upright column of the same height (the base
            areas being equal).

             Which is Heavier?
            On one pan of scales is placed a pail that is filled to the
            brim with water. On the other pan, exactly the same
            sized pail is placed, also brimful, but with a piece of
            wood floating in it (Fig. 87). Which pail will be
Figure 87

                I asked various people this question and got
             conflicting answers. Some answered that the pail with
             the wood would be heavier because "the pail has the
             water and the wood." Others held that, on the contrary,
             the first pail would be heavier "since water is heavier
             than wood."
                Both views are a mistake for both pails have the
             same weight. True, there is less water in the second pail
             than in the first because the floating piece of wood
             displaces some water. The immersed part of every
            floating body displaces exactly the same weight of water
             as the whole of the body weighs. That's why the scales
             will be in equilibrium.
                Another problem. Suppose I place on the scales
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            a glass of water and put a weight near it. When the
            system is balanced by the weights on the other pan,
            I drop a weight into the glass. What will happen with
            the balance?
              According to the Archimedes principle the weight in
            the water becomes lighter than before. It might be
            expected that the pan with the glass would rise but in
            actual fact the scales will remain in equilibrium.
              The weight in the glass has displaced some water,
            which has risen above the initial level, with the result
            that the pressure on the bottom of the vessel has
            increased so that the bottom is acted upon by an added
            force equal to the weight lost by the weight.

            Water on a Screen
            It turns out that water can be carried on a screen in
            real life and not only in fairy tales. A knowledge of
            physics will help to make this proverbially impossible
            thing possible. You'll need a wire screen about 15
            centimetres across with a mesh size of about
            1 millimetre. Immerse the network into melted wax and
            when it is taken out of the wax the wire will be covered
            with a thin layer of wax hardly noticeable for the naked
               The screen will still remain a screen - a pin will freely
            pass through its m e s h - b u t now you will be able
            literally to carry water on it. The screen will hold
            a fairly high level of water without any seepage through
            the mesh. You need only to pour the water carefully
            and see to it that the screen is not jerked.
               Why then doesn't the water seep? Because it doesn't
            wet wax and thus forms thin films between the meshes
            and it is the films' downward convexity that holds the
Figure 88   water (Fig. 88).
               Such a waxed screen may be placed on water and it
            will remain on the surface. It is thus possible not only
            to carry water on a screen but to float on it too.
               This paradoxical experiment accounts for a number
            of the everyday phenomena we take for granted
            because we get used to them so. Tarring barrels and
            boats, painting with oil paints and, in general, coating
             things we want to render water-tight with oily
            materials, and the rubberizing of fabrics, these are all
             nothing but the making of "screens" like the one just
             described. The idea behind each phenomenon is the
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same. Only in the case of the screen it appears in
a somewhat unusual disguise.

Soap Bubbles
Can you make soap bubbles? This is not as simple as it
might seem. I also once thought that it didn't take
much dexterity until I found out practically that
blowing large and beautiful bubbles is an art that
requires much exercise.
   But is it worthwhile to occupy yourself with such
a trifling business as blowing soap bubbles? Used as
a figure of speech the notion of soap bubbles is not
complimentary. But the physicist has another view of
them. The great English scientist Lord Kelvin wrote,
"Blow a soap bubble and observe it, it may take
a lifetime to investigate it, incessantly deriving lessons
of physics from it."
   Indeed, the fabulous play of colours on the surface of
thin soap films enables physicists to measure the
wavelengths of light, and the study of the tension in
these frail films gives an insight into the laws governing
the interaction between particles, those cohesion forces
without which there would be nothing in the world but
fine dust.
   The several experiments that follow do not pursue
such serious objectives, they are just amusements that
will only acquaint you with the art of blowing soap
bubbles. In his book Soap Bubbles the English physicist
Charles Boys gave a detailed account of a number of
experiments involving them. Those interested are
referred to this fascinating book, but we'll only describe
the simplest of the experiments here.
   These can be performed using a solution of
a conventional soap*, but for best results olive- or
almond-oil soaps are recommended. A piece of soap is
carefully dissolved in pure cold water until a fairly
thick solution is obtained. Rain or thaw water is the
best but if it's unavailable cooled boiled water will do.
For bubbles to have a long life it is recommended to
add one third of the volume of glycerin. Using a spoon,
remove foam and bubbles from the surface and insert
into it a long clay tube whose end on the outside and
inside has already been smeared with soap. Good
results are also achieved with straws about 10

   * Toilet soaps are unsuitable.
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            centimetres long that are split across the end.
               The bubbles are blown thus: dip the tube into the
            solution holding it upright so that a liquid film be
            formed at the end and carefully blow into it. Since the
            bubble is filled with warm air from your lungs, which is
            lighter than the surrounding air in the room, the
            bubble just blown will rise into the air.
               If from the very beginning you can produce a bubble
            10 centimetres in diameter, the solution is good,
            otherwise some more soap will have to be added to the
            liquid until bubbles of the above-mentioned size are
            obtained. But this test is not sufficient. When a bubble
            is produced, dip a finger into the soap solution and try
            to punch the bubble. If it doesn't burst you may
            proceed to the experiments, but if the bubble doesn't
            survive the test, add some more soap.
               Experiments should be carried out carefully, slowly,
            and quietly. If possible, the illumination should be
            bright, otherwise the bubbles will not show their
            iridescent play.
               The following are a number of entertaining
            experiments with soap bubbles.
                A Bubble Around a Flower. Pour some soap solution
            onto a plate or a tray so that the bottom is covered
            with a layer 2-3 mm thick. Place a flower or small vase
            in the middle and cover it with a glass funnel. Then,
            slowly lifting the funnel, blow into the narrow tube to
             form a soap bubble. Once the bubble has reached
             a largish size, tip the funnel over as shown in Fig. 89,
Figure 89    and liberate the bubble from under it. The flower will
             then be under a transparent hemispherical hood of
             soap film which will show all the colours of the
                Instead of a flower you can take a small statue and
            crown its head with a soap bubble. First you need to
            drop some solution onto the head of the statue and
            then, after blowing the large bubble, pierce it and blow
            a smaller one inside it.
               Bubbles Inside One Another. Blow a large bubble
            using the funnel, then immerse a straw into the soap
            solution so that only the end you will put into your
            mouth is dry and poke it carefully through the wall of
            the first bubble to the centre. By carefully drawing the
            straw back, a second bubble can be blown inside the
            first one, then a third, fourth, and so on.
               A Cylinder of Soap Film (Fig. 90) can be blown
            between two wire rings. In order to do this lower
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            a conventional ball-shaped bubble onto the bottom
            ring. Then put a wetted second ring over the top of the
            bubble and by raising it the bubble will extend until it
            becomes cylindrical. Curiously enough, if you raise the
Figure 90   upper ring higher than the length of the ring's circum-
            ference, the cylinder will become narrower at one end
            and wider at the other, and then it will disintegrate into
            two bubbles.
               The film of a soap bubble is always in tension and
            exerts a pressure on the air inside it. By directing the
            funnel at the flame of a candle you can make sure that
            the force of the thin film is not all that negligible since
            the flame will be deflected quite a bit (Fig. 90).
               It is interesting just to observe a bubble when it is
            taken from a warm room into a cold one. It will shrink
            appreciably, and conversely it will expand when
            brought from the cold room into the warm one.
            Clearly, the reason is that the air in the bubble expands
            and contracts. If, for example, the volume of a bubble
            at — 15 °C is 1,000 cubic centimetres and the bubble is
            brought into a room at + 1 5 ° C , its volume will
            increase by about 1,000x 30 x 1/273 or about 110
            cubic centimetres.
               Furthermore, it should be noted that the common
            idea that soap bubbles are short-lived is wrong since
            with adequate handling a soap bubble can survive for
            weeks. The English physicist Dewar (famous for his
            works on air liquefaction) kept soap bubbles in special
            bottles that were protected from dust, drying and jerks.
            Under these conditions he managed to keep some
            bubbles for a month or so. An American, Lawrence,
            succeeded in keeping soap bubbles in a glass cup for

            An Improved Funnel
            Those who have poured water through a funnel into
            a bottle know that it is necessary to raise the funnel
            from time to time, otherwise the liquid will not pour
            out of it. It's the air inside the bottle that, when
            compressed by the incoming liquid and unable to
            escape, stops more liquid from coming in from the
            funnel. Understandably, by raising the funnel we let the
            compressed air out, thus again enabling some more
            liquid to go in.
               It would perhaps be quite practical to design a funnel
            so that it has longitudinal crests on its outer surface
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            that would keep the funnel from sticking to the
            bottleneck. But I haven't ever seen such a funnel in
            everyday life, only in laboratories they use a filter
            designed after this fashion.

            How Much Does Water Weigh in
            a Glass Held Upside-Down?
            You'd say, "Nothing, of course. Water won't stay in the
               I'd ask, "And if it does stay, what then?"
               Actually, it is possible to keep water in a glass held
Figure 91   upside-down so that it doesn't pour out. The method is
            shown in Fig. 91. An upturned goblet tied at the
            bottom on one side of a balance is filled with water so
            that it doesn't pour out because the goblet's edges are
            immersed in water. On the other side of the balance tie
            an empty goblet, exactly the same sort.
              Which side will go down?
              That to which the upturned goblet with water is tied.
            This goblet is exposed to atmospheric pressure from
            above, and from b e l o w - t o the atmospheric pressure
            minus the weight of the water contained in the goblet.
            For the system to be in balance you'd have to fill the
            other goblet with water. Accordingly, the water in the
            upturned glass weighs in these circumstances as much
            as it would in a normally held glass.

            How Much Does the Air in
            a Room Weigh?
            Can you say, however inaccurately, how much the air
            in a small room weighs? Several grammes or several
            kilogrammes? Would you be able to lift such a load
            with a finger or would it be difficult to hold it on your
               Perhaps these days there is no one who believes air is
            weightless as was widely held in ancient times. But even
            today many wouldn't estimate its weight.
               Remember that a litre jar of the warm summer air
            near the ground (not in the mountains) weighs 1.2
            grammes. A cubic metre holds 1,000 litres and therefore
            weighs 1,000 times as much, i.e. 1.2 kilogrammes.
              Now we can easily work out the weight of the air in
            a room. To do so, we'll only need to know how many
            cubic metres there are in it. If, say, the area of the room
            is 15 square metres, the height is 3 metres, then it
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contains 15 x 3 = 45 cubic metres. The air thus weighs
45 kilogrammes plus 1/5 of 45, i.e. 9 kilogrammes,
which makes 54 kilogrammes in all. You could not
move this load with a finger or carry it about on your
back with ease.

An Unruly Cork
This experiment will vividly demonstrate               that
compressed air has a force and an appreciable one at
   For the experiment we'll only need a common bottle
and a cork that's somewhat smaller than the
   Hold the bottle horizontally, insert the cork into the
neck and ask somebody to blow the cork inside the
   No problem, it would seem. But try it, blow hard at
the cork, you'll be amazed at the result. The cork won't
be driven inside the bottle but... will fly into your face!
   The harder you blow the faster it'll shoot out.
   If you want the cork to slide inside you'll have to do
quite the opposite-not to blow at the cork but to suck
the air from the hole.
   These strange phenomena can be explained as
follows. When you blow into the bottleneck you drive
some air through the gap betwen the cork and the wall
of the neck. This increases the pressure inside the bottle
and throws the cork out. If then you suck the air out,
the air inside the bottle becomes thinner and the cork
is pushed inside by the pressure of the air outside. The
trick works out well only when the neck is absolutely
dry as a wet cork sticks.

The Fate of a Balloon

Balloons sometimes go astray. But where? How high
can they fly?
   A balloon that escapes is carried always not to the
boundaries of the atmosphere, but only to its "ceiling",
i.e. to a height where the air is thin and the weight of
the balloon equals that of the air displaced by it. But it
does not always reach its ceiling. Since it swells (due to
the reduction in the external pressure) it may burst
before it reaches the ceiling.
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            How to Blow Out a Candle?
            It's child's play, you might think, to blow out a candle.
            But occasionally an attempt is a failure. Try and blow
            a candle out through a funnel and you'll see that this
            requires especial dexterity.
               Place the funnel against the flame of a candle and
            blow at it through the thin end. The flame won't so
            much as stir, although the stream of air from the funnel
            would seem to be striking the flame directly.
               Perhaps you now think the funnel is too far away
            from the flame, and so you bring it nearer and again
Figure 92   begin to blow hard. You might be shocked by the
            result: the flame deflects not away from you but
            towards you, against the stream of the air coming from
            the funnel.
               What is to be done then to kill the candle flame? It
            is necessary to locate the funnel so that the flame is not
            on the axis of the funnel but in the line of its cone part.
            Now by blowing into the funnel you'll easily extinguish
            the candle.
               This is explained by the fact that the air stream
            leaving the narrow part of the funnel does not
            propagate along its axis but spreads along the walls of
            the cone, thus forming a sort of an air vortex. But the
            air along the funnel axis is rarefied with the result that
            a return air flow sets in near it. It is now clear why
            a flame located on the axis of the funnel leans towards
            the funnel, and when the flame is on the periphery of
            the cone, it bends the other way and goes out.

            A car wheel with a tyre is rolling to the right, its rim
            rotating clockwise. The question is: in what direction
            does the air inside the tyre move-against the direction
            of rotation or in the same direction?
              The air moves away from the place of compression in
            both directions - forwards and backwards.

            Why Are There Gaps Between the Rails?
            Railway builders always leave gaps between the butts of
            adjacent rails on purpose. Without the gaps the railway
            would soon fall into disrepair. The reason is that all
            things expand on heating. A steel rail, too, elongates in
            summer, heated by the sun. If no space were allowed
            for the rails to expand, these would push against
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adjacent rails with an enormous force and bend
sideways wrenching out the spikes and destroying the
   The gaps are designed with due account of winter
temperatures. In winter the rails shrink from cold,
thereby additionally increasing the gaps. Therefore, they
are calculated very carefully considering the local
   An example of the use of the property of a body to
shrink on cooling is the old procedure of shoeing cart
wheels. A heated iron shoe is slided onto the rim of
a cart wheel. When the shoe is allowed to cool down, it
shrinks and squeezes tightly onto the rim.

A Glass and Tumbler
You may have noticed that tumblers for cold drinks are
often made with a thick bottom. The reason is obvious:
such a tumbler is more stable. Why then don't we use
tumblers for hot drinks? After all it would be better for
glasses to be more stable in that case too.
   Thick-bottomed tumblers are not used for hot drinks
because the walls of such tumblers would be heated by
the hot liquid and expand more than the thick bottom.
The thinner the glassware and the less difference there
is between the thickness of the wall and the bottom, the
more uniform will be the heating and the less the risk
of cracking.

The Hole in the Cap of a        Tea-Kettle
The cap of a metallic tea-kettle has a hole. What for?
To let some vapour out, otherwise it will pop the cap
off. But the cap expands on heating in all directions.
What happens to the hole in the process? Does it
become narrower or wider?
  It becomes wider. In general the volume of holes and
cavities becomes larger on heating in exactly the same
way as an equal piece of surrounding material does.
For that reason, by the way, the capacity of vessels
increases on heating, not decreases as is widely

Why does smoke go up in still weather?
 The smoke from a chimney ascends because it's
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          carried by hot air that expands on heating, thus
          becoming lighter than the air around the chimney.
          When the air supporting the smoke particles cools
          down, the smoke descends and spreads out over the

          Incombustible     Paper
          We can perform an experiment in which a paper strip
          doesn't burn in the flame of a candle.
             Wind a narrow paper strip tightly around an iron
          rod. If now you introduce the rod with the wound strip
          into the flame of a candle, the paper won't catch fire.
          The fire will lick the paper, the latter will char but not
          burn down until the rod becomes hot.
             Why? Because iron, just like any metal, is a good
          heat conductor; it leads away the heat obtained by the
          paper from the flames. Replace the metal rod by
          a wooden stick and the paper will burn because wood
          is a poor heat conductor. With a copper rod the
          experiment is even more successful.
             Instead of the paper strip you could also use a piece
          of string wound tightly around a key.

          How to Seal Window Frames for Winter
          An adequately sealed window frame keeps out cold.
          But to seal it properly you should get it right why the
          frame "heats" a room.
             Many believe that a second frame is used in winter
          because two windows are better than one. That is not
          so. It's not the second window that matters here but
          the air confined between the windows.
             Air is a very poor heat conductor. Therefore, some
          air adequately confined for it not to carry any heat
          away prevents the room from cooling.
             But for best results the air must be sealed tightly
          inside. Some people wrongly think that when a frame is
          sealed for the winter the upper gap in the external
          frame should be left unsealed. Should you do so the air
          within the cavity would be displaced by outside cold
          air, thus chilling the room. On the contrary, both
          frames should be treated painstakingly and not even
          the tiniest chink should be left.
             Alternatively, you can with good results glue frames
          over with strips of strong paper. Well sealed or glued
          windows cut down your heating expenses.
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Draught from a Closed           Window
It might seem unusual that in a cold weather there is
often a draught from a window that is tightly closed,
carefully sealed and does not have the smallest hole.
There is nothing surprising about that.
   The air inside a room is almost never at rest. There
are invisible flows caused by the heating and cooling of
the air. Heating makes air thinner, and hence lighter.
Conversely, cooling makes it denser and heavier. The
light, heated air over a lamp or stove is displaced by
cold air up to the ceiling because the heavy air that has
cooled near the windows or cold walls, flows down to
the floor.
   These currents in a room are readily discovered using
a balloon with a small weight attached to it for it not
to strike the ceiling and float freely in the air. Let the
balloon go near a warm stove and it'll travel about the
room pulled around by the invisible air currents: from
the stove to the window under the ceiling, then down
to the floor and back to the stove for a new cycle.
   That's why in winter we feel a draught from
a window, especially at the bottom, even though the
frame is securely sealed and keeps the outside air out.

How to Chill with Ice
If you want to chill a bottle of drink, where should you
place it, on or under the ice?
   Many put a bottle on the ice without a moment's
hesitation, just like they put a tea kettle on a fire.
That's not the way to do it. Heating should be done
from below, but chilling, on the contrary, is better from
the top.
   Explain why. You know that colder substances are
denser than warm ones. Thus a chilled beverage is
denser than a warm one. When you place the ice over
the top of the bottle, the upper portions of the drink
(adjacent to the ice) sink on cooling being replaced by
another amount of the liquid that in turn cools down
and descends as well. In a short while all the drink in
the bottle will have been in contact with the ice and
chilled. But if the bottle is placed over the ice, the
lowest portion cools first, its density increases and it
stays at the bottom making no room for the rest of the
liquid that is warmer. No mixing occurs here and the
chilling is extremely poor.
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             It pays to chill everything from the top and not just
          drinks-meat, vegetables and fish should be placed
          under ice as well. They are chilled not so much by the
          ice itself as by the surrounding air because the cold air
          comes down. If you need to cool a room with ice don't
          place it on the floor but put it up high on a shelf or
          suspend from the ceiling.

          The Colour of Water Vapour
          Have you ever seen water vapour? Could you say what
          colour it is?
             Strictly speaking, water vapour is absolutely
          transparent and colourless. It is invisible, just like air.
          The white fog that is popularly known as "vapour" is
          really a multitude of water droplets, it is a suspension
          of fine water particles, not vapour.

          Why Does a Boiler "Sing"?
          A boiler or a kettle produces a singing sound when the
          water is about to boil. The water adjacent to the heater
          vaporizes to form small bubbles. Being much lighter
          these are expelled upwards by the surrounding water
          and as they go up the bubbles pass through water that
          has a temperature of less than 100 °C. The vapour in
          the bubbles cools, contracts and the bubbles collapse
          under pressure. Thus, just before boiling sets in, more
          and more bubbles go up but fail to reach the surface
          collapsing on the way to produce a cracking sound. It
          is these numerous cracking that produce the sound we
          hear at the outset of boiling.
             When the water eventually heats to boiling
          temperature, the bubbles cease to collapse on their way
          up and the "singing" discontinues. However, once the
          water starts to cool down, again the earlier conditions
          occur and the "singing" resumes.

          A Miraculous      Top
          Cut a small square out of thin tissue-paper. Fold it
          diagonally twice and smooth it out again. You'll thus
          know where the centre of mass of the square is. Now
          place the paper on the point of an upright needle so
          that the latter supports it at the middle.
            The paper will balance since it's supported at the
          centre of mass. A slightest flow of air will make it
Figure 93
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          in the coat and allow the other to stand in the room
          uncovered. When the ice in the uncovered bottle has
          melted, unfold the fur coat and you'll see that the ice is
          nearly intact. In consequence, the fur coat not only
          didn't heat the ice, but, as it were, cooled it, thus
          hampering 4he melting!...
             What objections could be raised here? How could
          you refute the arguments?
             There is no objecting or refuting. In fact, fur coats
          don't heat things up if by "heat up" we mean to impart
          heat. A lamp heats, a stove heats, a human body heats,
          too, because all of these bodies are sources of heat. But
          a fur coat is not. It generates no heat, but only stops the
          heat of our body from going astray. That's why
          a warm-blooded animal whose body itself is a source of
          heat will be warmer with a fur coat than without it. But
          the thermometer generates no heat of its own and its
          temperature won't change in the coat. The ice in the
          coat retains its low temperature longer because the fur
          c o a t - a fairly poor heat conductor-hinders the passage
          of heat from the outside.
             Snow "heats" the earth in the same way a fur coat
          does. A loose powder substance, snow is a poor heat
          conductor and helps to keep cold out. Not infrequently
          a thermometer in snow-covered soil indicates it is as
          much as ten degrees hotter than is exposed soil.
          Farmers are well aware of this heating effect of a snow
             Thus, the answer to the question of whether a fur
          coat heats or not is that it only helps us to heat
          ourselves. Or rather we heat the fur coat, not vice

          How to Air Rooms in Winter
          The best way to air a room is to open a window when
          a fire is burning. Fresh, cold outside air will then force
          out the warm, lighter air from the room into the
          fire-place and out through the chimney into the
             However, do not think that the same thing will occur
          when the window is closed, for the outside air will leak
          into the room through gaps in the window, walls, etc.
          True, some of it will really get into the room but not
          enough to sustain the fire. Therefore, apart from the
          outside air some air must come from other rooms
          where it might be neither pure nor fresh.
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Figure 94

Figure 95

               The two accompanying figures demonstrate the
            difference between the two cases. The arrows indicate
            the flow of air.

            Where to Arrange a Ventilation Pane
            Where? At the top or bottom of a window? In some
            homes ventilation panes are at the bottom. Admittedly,
            these are convenient to open and close, but they are
            inefficient. Let's consider the physics of the air exchange
            through the ventilation pane. Outside air is colder than
            that inside and displaces the latter. However, it
            occupies the part of the room below the ventilation
            pane. The air above the pane doesn't contribute to the
            exchange, i.e. is not ventilated.

            Paper Saucepan
Figure 96
            Look at Fig. 96. An egg is being boiled in a paper
               You'd say, "Oh, but the paper'll now catch fire and
            the water will pour out!"
               Try the experiment on your own. Make the
            "saucepan" from parchment paper and attach it to
            a wire holder. The paper won't be destroyed by the fire.
            The reason is that water in an open vessel can only be
            heated to boiling temperature, i.e. 100°C. Therefore,
            the water, which has a large thermal capacity, absorbs
            excess heat from the paper and so does not allow it to
            heat up more above 100° to a point when it might
            ignite. (Perhaps it would be more convenient to make
            use of a small paper box as shown in the figure.) So the
            paper does not catch fire although flames touch it.
              A similar kind, but disastrous, "experiment" is at
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          times performed by absent-minded people who put an
          empty kettle onto a fire with the pitiful result that the
          latter gets unsoldered. The reason is clear now: solder
          is relatively low-melting and it is only its close contact
          with water that saves it from its temperature. rising
          dangerously. This applies to all sorts of soldered things.
             Further, you could melt a piece of lead in a small
          box made of a playing card. You'll only need to expose
          to flames the place that is in direct contact with the
          lead. Being a good heat conductor, the metal will
          quickly take away heat from the paper. The
          temperature of the paper will thus be maintained at
          about 335 °C (melting point for lead), which is
          insufficient to ignite the paper.

          What is the Lamp Glass for?
          Few people know what a long history the lamp glass
          went through before it appeared in its present-day
          form. For millennia people had used flames for lighting
          without resorting to the services of glass. It took the
          genius of Leonardo da Vinci (1452-1519) to introduce
          this important improvement of the lamp. But Leonardo
          used a metal tube, rather than a glass one, to surround
          the flame. Three more centuries passed before the metal
          tube was replaced by the transparent cylinder. You see
          thus that the lamp glass is an invention developed by
          scores of generations.
             What's its purpose?
              Not all of you will come up with the right answer to
          this natural question.
              To protect the flames from wind is only a secondary
          role of the glass. Its main effect is to increase the
          brightness of the flames, to boost the combustion
          process. The role of the glass here is like that of
          a chimney or stack; it intensifies the inflow of air to the
          flames, i.e. improves the "draught".
              Let's take a closer look at this. The column of air
          inside the glass is heated by the flames much faster
          than the air surrounding the lamp. After it has heated
          and thereby become lighter, the air is displaced
          upwards by the heavier cold air arriving from below
           through holes in the burner. This results in a steady
           flow of air upwards, a flow that continually takes the
           combustion products out and brings fresh air in. The
           higher the glass, the more difference there is between
           the heated and unheated air columns and the more
1 — 975
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intensive is the inflow of fresh air, and hence the burn-
ing. The situation is like that in industrial chimney
stacks which is why they are made so high.
   Interestingly, even Leonardo had understood these
phenomena. In his manuscripts he says, "Where fire
appears, an air flow forms around it, the flow supports
and intensifies it."

Why Doesn't a Flame Go Out by Itself?
A closer examination of the process of combustion
inevitably leads to the above question. After all, the
combustion products are carbon dioxide and water
vapour, noncombustible       substances incapable     of
supporting the process. Accordingly, once started
a flame must be surrounded by noncombustibles that
hinder the inflow of air. Combustion cannot occur
without air and the flame would be bound to die out.
   Why then this is not the case? Why does the process
of combustion carry on as long as there is a supply of
combustibles? For the only reason that gases expand
on heating and become lighter. It's owing to this that
heated combustion products don't stay where they've
been formed, i.e. in the immediate neighbourhood of
the flames, but are at once forced upwards by fresh air.
If the principle of Archimedes didn't apply to gases (or
there were no gravity), any flame would go out after
a while on its own.
   You can easily verify that combustion products kill
a flame. At times you make use of this unawares to
extinguish the fire in a lamp. How do you blow out
a kerosene lamp? Blow into it from above, i.e. do not
let the combustion products out. The flames go out
deprived of the supply of fresh air.

Why Does Water Kill Fire?
A seemingly simple question... that is not always
correctly answered.
   Let's briefly explain the phenomenon.
   First, on touching a hot body water turns into
vapour, so taking heat away from the burning body. To
convert boiling water into vapour takes more than five
times as much heat as is required to heat the same
amount of cold water to 100 °C.
   Second, the resulting vapour occupies hundreds of
times more space than the source water. The vapour
130-131   Seventy-Five More  Questions
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          envelopes the body, cutting off the air that is
          indispensable for its burning.
            To improve the fire-extinguishing power of water
          they sometimes add ... gunpowder to it. Strange as it
          might seem, the measure is quite reasonable because
          the powder burns down quickly evolving a great
          amount of noncombustible gases that cover the burning
          material to hinder combustion.

          Heating with Ice and Boiling Water
          Is it possible to use a piece of ice to heat another? Or,
          to cool?
             Is it possible to heat one quantity of boiling water
          with another?
             If some ice at a low temperature, — 20 °C say, is
          brought into contact with a piece of ice at a higher
          temperature, — 5 °C say, then the first piece of ice will
          heat up (become less cold), and the second will cool
          down. Therefore, it is quite possible to cool or heat ice
          with ice.
             But one body of boiling water cannot heat another
          body of boiling water (at the same pressure), for at
          a given pressure boiling water is always at the same

          Can You Bring Some Water
          to the Boil Using Other Boiling Water?
          Pour some water into a small bottle (jar or phial) and
          place it in a saucepan with pure water so that it doesn't
          touch the bottom. Of course, you'll have to suspend the
          bottle from a piece of wire. Put the saucepan on a fire.
          When the water in the saucepan boils, it would seem
          that the water in the bottle should also boil shortly.
          Only you will never see this, however long you wait.
          The water in the bottle will get hot, very hot, but boil
          will it not. The boiling water appears to be too cool to
          bring another body of water to the boil.
             Quite an unexpected result, it seems, but let's analyse
          it more closely. To bring water to the boil it is not
          sufficient only to heat it to 100°C-it also needs
          a substantial supply of so-called latent heat. Pure water
          boils at 100 °C, and under standard conditions its
          temperature never exceeds this, however long you heat
          it. In our case the source of heat used to heat the water
          in the bottle has a temperature of 100 °C. It, too, is
Figure 97

Figure 98
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Accordingly, material absorbing the fat should        be
placed at the side opposite to the iron.

How Far Can You See From High Places?
From a flat place we only see the group up to a certain
boundary. This boundary of view is called the "horizon
line". Trees, houses and other high structures lying
beyond the horizon line are seen not in full, because
their lower parts are blotted by the convexity of the
earth. Even plains or the sea, although apparently flat,
are in fact convex, for they are parts of the curved
surface of the globe.
   How far then does an average-sized man see over
a plain?
   He can only see up to 5 kilometres. To see beyond
that he'll have to climb up higher. A man on horseback
on a plainland would see up to 6 kilometres and
a sailor on a mast 20 metres high would see the sea
around him up to 16 kilometres away. From the top of
a lighthouse towering above water at 60 metres the sea
is seen for nearly 30 kilometres.
   But, of course, the widest panoramas open up before
airmen. From an altitude of 1 kilometre they can see
almost for 120 kilometres in all directions, if not
hindered by clouds or fog. At twice the height an
airman will see for 160 kilometres using a perfect
optical device. Further, from 10 kilometres one can see
within 380 kilometres, and astronauts orbiting the
Earth see the whole of one side of the globe.

Where Does a Chirring Grass Hopper Sit?
Sit somebody in the middle of a room, with his eyes
blindfolded, and ask him to sit still and not turn his
head. Take then two coins and tap one on the other at
various places in the room but at about the same
distance from your friend's ears. Ask your friend to
guess the place whence the sound comes. It will be diffi-
cult to do and your friend will point in some other
   If you step aside, the errors won't be as bad because
now the sound in the nearest ear of your friend will be
heard somewhat louder, so enabling him to determine
the location of the source.
   The experiment makes it clear why it's impossible to
spot a grass hopper chirring in the grass. The sharp
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          sound is heard two paces away from you. You look
          there and see nothing, but now the sound is distinctly
          heard from the left. You turn your head in that
          direction, but no sooner have you done that than the
          sound already comes from some other direction. The
          speed of the grass hopper stuns you and the quicker
          you turn to the direction of the singing insect the
          quicker the invisible musician hops about. But in
          reality the insect is sitting placidly in place and his
          "hops" are just an illusion. Your problem is that when
          you turn your head you put it exactly so that the grass
          hopper becomes equally separated from both of your
          ears. This condition (as you should know it from the
          experiment just described) is conducive to an error. If
          the chirring comes from ahead of you, you place it,
          erroneously, in the opposite direction.
             In consequence, if you want to determine where
          a sound comes from, you should not turn your head
          towards the sound, but conversely, turn it away. Which
          is exactly what we do when we, as it were, "prick up
          our ears".

          When a sound we have produced is reflected from
          a wall or another obstacle and returns to our ears, we
          hear an echo. It is only heard distinctly if the time-lag
          between the sound generation and its return is not too
          short. Otherwise the reflected sound would melt with
          the initial one and amplify it, the sound will then
          reverberate, e.g. in large empty halls.
             Imagine that you are standing in an open place and
          there is a house in front of you 33 metres away. Clap
          your hands. The sound will travel through the 33
          metres, reflect from the walls and come back. How long
          will that take? Since the sound covered 33 metres there
          and the same distance back, it'll return in 66/330 or 1/5
          of a second. Our sharp sound was so short that it
          terminated in less than 1/5 second, i.e. before the echo
          arrived. The two sounds didn't merge and were heard
          separately. A monosyllabic word ("yes", "no", etc.) is
          pronounced in about 1/5 second and we can hear such
          words echoed at a distance of only 33 metres from an
          obstacle. But for bisyllabic words the echo merges with
          the initial sound intensifying it but rendering it obscure,
          we don't hear it separately.
             At what distance must the obstacle be then so that
Figure 99
136-137   Seventy-Five More  Questions
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          To See Through a Palm
          Fold a sheet of paper into a tube, bring it up to your
          left eye with your left hand and look through it at some
          distant object. Now bring your right palm near to your
          right eye so that it nearly touches the tube. Both hands
          should be about 15-20 centimetres away from the eyes.
          You'll then make sure that your right eye sees perfectly
          through your palm as if there were a round hole in it.
             The reason of this unexpected phenomenon was as
          follows. Your left eye prepared to view a distant object
          through the tube and the crystalline lens adapted
          accordingly. The eyes function in such a way that they
          always adapt in sympathy.
             In the experiment described the right eye, too,
          adapted to distant- sight with the result that the near
          palm appeared blurred to it. In short, the left eye
          clearly sees the distant object, the right one, sees the
          palm unclearly. The net result is that it seems to you
          that the distant object is seen through the shielding

          Through    Binoculars
          At a seaside you are watching a boat approaching the
          shore through a pair of binoculars that magnifies three
          times. How many times will the speed be increased with
          which the boat is approaching the shore?
             Assume that the boat is sighted 600 metres away and
          is approaching the observer with a speed of 5 metres
          per second. Through binoculars with triple mag-
          nification the boat at 600 metres appears to be at 200
          metres. A minute later it will be 5 x 60 = 300 metres
          closer and will then be 300 metres away from the
          observer. In the binoculars its apparent size would
          indicate it were 100 metres away. Consequently, an
          observer looking through the binoculars would think
          the boat has travelled 200 — 100 = 100 metres, whereas
          in actual fact it has actually covered 300 metres. It
          follows that the speed at which the boat approaches
          when observed through the binoculars has decreased
          not increased by three times.
             The reader can arrive at the same result by another
          argument, i.e. by taking the initial distance, speed and
             The speed with which the boat approaches has thus
          reduced by as many times as the binocular magnifies.
Seventy-Five More   Questions
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From the Front or the Back?
There are many things in each household that are used
inefficiently. I've already mentioned that some people
cannot use ice properly to chill drinks-they place them
on the ice instead of under it. It appears that some
people cannot use a conventional mirror either. Quite
frequently, if they want to see themselves better in the
mirror they turn the light on behind themselves in order
to "illuminate the reflection", instead of illuminating
themselves from the front.

Drawing Before the Mirror
That a mirror reflection is not identical with the
original may be demonstrated by the following
   Stand or hang an upright mirror in front of you on
the table, place a sheet of paper on it and try to draw
something, for example, a rectangle with diagonals. But
in doing so don't look directly at your hand, but follow
the movements of its reflection in the mirror.
   You'll find that this seemingly simple problem is
almost intractable. Over the years our visual
perceptions and motions have been correlated but the
mirror violates this and represents our motions to our
eyes in an inverted form. Long-term habits rebel
against each our motion: you want to draw a line to
the right, say, but the hand draws to the left, and so on.
   Stranger things will occur if instead of a simple figure
you attempt to draw more intricate designs or write
something whilst looking in the mirror. The result will
be a funny confusion.
   The impressions left on carbon paper are inverted
'ettering, too. Just try to read the text on it. Quite
a challenge! But bring it to a mirror and the text will
appear in its habitual form. The mirror gives the
reflection of what is itself an inverted image of normal

Black Velvet and White Snow
Which is the lighter - black velvet on a sunny day or
pure snow on a moonlit night?
  Nothing, it seems, surpasses black velvet in blackness
or white snow in whiteness. These age-old metaphors of
white and black appear, however, quite different when
viewed by a physical instrument-a photometer. It then
138-139   Seventy-Five More Questions
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          turns out that the blackest velvet in sunlight is lighter
          than the purest snow in moonlight.
             This is because a black surface, however dark it
          might be, doesn't completely absorb all the visible
          incident light. Even soot and platinum black-the
          blackest substances known-scatter about 1-2 per cent
          of the incident light. We take 1 per cent for argument's
          sake and suppose that snow scatters 100 per cent of the
          incident      light    (which    is    undoubtedly      an
          overstatement)*. It is known that the illumination
          provided by the sun is 400,000 times that of the moon.
          Therefore, the 1 per cent of sunlight scattered by the
          black velvet is thousands of times more intense than
          the 100 per cent of moonlight scattered by snow. In
          other words, sunlit black velvet is many times lighter
          than moonlit snow.
             To be sure, this is true not only of snow but also of
          the best white pigments (the whitest of them
          all - lithopone - scatters 91 per cent of light). Since no
          surface, unless it's hot, can beam out more light than
          strikes it, and the sun sends out 400,000 times as much
          light as the moon, it's impossible to have a white
          pigment that would in moonlight be lighter than the
          blackest pigment on a sunny day.

          Why is Snow White?
          Why, indeed? It consists of transparent ice crystals.
             For exactly the same reason that ground glass and
          all ground transparent substances in general are white.
          Grind some ice up in a mortar or chip it with a knife
          and you'll get white powder. The colour is due to the
          fact that light, when penetrating into tiny pieces of
          transparent ice, doesn't pass through them but reflects
          inside them at the boundaries between the ice and the
          air (total internal reflection). But a randomly scattering
          surface is perceived by the eye as white.
             Thus, snow is white because it consists of tiny
          particles. If the gaps between the snow flakes are filled
          with water, the snow becomes transparent. Such an
          experiment is easy. Put some snow into a jar and pour
          some water into it, and before your very eyes the snow
          will become colourless, transparent.

             * Fresh snow only scatters about 80 per cent of light.
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The Shine on a Blackened        Shoe
Why does a blackened shoe shine?
   Neither the sticky black shoe polish, nor the brush
seem to have anything to impart the gloss to shoes.
Therefore, it's a mystery for many.
   We'll first clear up the difference between the glossy
polished surface and the dull one. It's widely believed
that the polished surface is smooth and the dull one is
irregular. This is not so: both are irregular. There are
no absolutely smooth surfaces. Examined uncier the
microscope polished surfaces would be like razor blades
and for a man reduced 10,000,000 times the surface of
a smoothly polished blade would appear to be a hilly
terrain. There are irregularities, depressions and
scratches on any surface, both dull and polished. What
matters is the size of these irregularities. If they are
smaller than the wavelength of the incident light, then
the rays are reflected correctly, i.e. at the angle of
incidence. Such a surface gives mirror reflections, it
shines and we call it polished. If, on the other hand, the
irregularities are larger than the wavelength of the
incident light, the surface scatters the ray randomly and
does not follow the reflection law. Such scattered light
gives no mirror reflections and highlights, and the
surface is called dull.
   This suggests, by the way, that a surface may be
polished for some rays and dull for others. For visible
light with a mean wavelength of about half
a micrometre (0.0005 mm) a surface with irregularities
of about that size will be polished; for infrared light,
which has longer wavelength it's polished, too. But
for ultraviolet light, which has shorter wavelength, it's
   But back to the pedestrian subject of our problem.
Why do polished shoes shine, after all? The
unblackened surface of leather has a highly irregular
microstructure with "peaks" larger than the mean
wavelength of visible light, it's dull. By blackening it we
smooth out the surface and lay the hairs that stick out
down. Brushing removes any excess polish at
projections and fills the troughs, reducing the
irregularities down to a size at which the peaks become
smaller than the wavelengths of visible rays and the
surface turns into a glossy one.
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          Through Stained Glass
          What colour are red or blue flowers viewed through
          green glass?
             Green glass will only transmit green light and catch
          all the rest. Red flowers send out mostly red light. If we
          look through green glass at a red flower we'll receive
          no light from its petals as the only rays it emits are
          retained by the glass. The red flower will therefore
          appear to be black through such glass.
             Now you should easily see that the blue flower
          viewed through green glass will be black as well.
             Professor M. Yu. Piotrovsky, a physicist, artist and
          acute observer of nature, made a number of interesting
          observations in his book Physics on Summer Outings.
             "Observing flowerbeds through a red glass we see
          that purely red flowers, say, geranium, appear to us as
          bright as purely white one; green foliage appears as
          absolutely black with a metallic lustre; blue flowers
          (aconite, etc.) are so black as to be next to impossible
          to make them out against the black background of the
          leaves and yellow, pink, and lilac flowers appear more
          or less dull.
             "Through a green glass we see the unusually bright
          green of the foliage and white flowers come out still
          more distinctly against it; somewhat more pale are
          yellow and blue ones; red flowers are jet black; lilac
          and light-pink colours appear as dull and grey so that,
          for example, the light-pink petals of a wild rose are
          darker than its richly coloured leaves.
             "Finally a blue glass will again make red flowers look
          black, white flowers will look bright, yellow-absolutely
          black, and blue and dark-blue-almost as bright as the
          white ones.
             "It's easily seen from this that red flowers really do
          emit much more red light than any other colour, yellow
          flowers emit about an equal amount of red and green,
          but very little blue, whilst pink and purple flowers emit
          a lot of red and blue, but very little green light."

          A Red Signal
          On the railways the stop signal is a red colour. Why?
            Red light has the longest wavelength in the visible
          spectrum and is thus less scattered by any particles
          suspended in the air than are other colours. Therefore,
          red light penetrates farther. It is of paramount
          importance to obtain the greatest visibility possible for
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a transport signal since to be able to stop his train the
engine-driver should begin breaking long before
reaching an obstacle.
  The greater transparency of the atmosphere to longer
waves, by the way, explains why astronomers use
infrared filters to photograph planets (especially Mars).
Fine details blurred in a conventional picture come out
distinctly on a photograph taken through a glass that
only transmits infrared light. In the case of Mars it's
possible to photograph the surface of the planet, while
a conventional picture only shows its atmospheric
   A further reason for selecting the red light for the
stop signal is that our eyes are more sensitive to this
colour than to blue or green.
142-143   Optical Illusions

          Tricks of Vision
          The optical, or visual, illusions to which this section is
          devoted are not accidental companions of our
          vision-they occur in definite circumstances, are
          governed by physical laws and affect any normal
          human eye. That human beings are subject to visual
          illusions and can be mistaken as to the source of their
          visual perceptions, should by no means be considered
          an undesirable disadvantage or an unqualified flaw in
          our constitution, whose removal would benefit us in
          many respects. The artist would rebel against such an
          "infallible" vision. For him our ability, under certain
          conditions, to see what really is not is a blessing
          enriching enormously the potentialities of the fine arts.
          The 18th century mathematician Euler wrote: "Artists
          are especially skilled at using this common illusory
          experience. The whole of the art of painting is based on
          this. If we were used to judge about things as these are
          in reality then this art would be impossible, it would be
          as if we were all blind. In vain the artist would exhaust
          his skill in colour blending, for we would merely say:
          there is a red spot on this board, here a blue one, here
          a black and there, several whitish lines. Everything is in
          the same plane and there is no difference in distance. It
          would thus be impossible to represent anything. No
          matter what were painted in the picture, it could seem
          to be like writing on a paper, and perhaps we would, in
          addition, try to make out the signification of all the
          coloured spots. For all the perfection, weren't we to be
          pitied greatly, being devoid of the pleasure we derive
          every day from such pleasant and useful arts!"
             Since the subject is of such lively interest for the
          artist, physicist, physiologist, physician, psychologist,
          philosopher, and for any inquisitive mind, many books
          and articles have been published in this country and
          elsewhere. *
             We'll here consider several types of tricks played by
          our unaided eye, i.e. without any appliances such as
          stereoscopes, punched cards, and so on.
             As to the causes of one or another visual illusion,
          only a relatively small number have well established,

               * See, e.g. The Nature of Experience (1959) by R. Brain; Optical
          Illusions (1964) by S. Tolansky; The Neurophysiological Aspects of
          Hallucinations and Illusory Experience (1960) by W. Grey.
Optical   Illusions

unquestionable explanations. These include those due
to the structure of our eyes, irradiation, Mariotte's
illusion (blind spot), astigmatism illusions, and so forth.
   As an instructive example we'll consider the optical
illusion of Fig. 141: white circles arranged in a certain
way on a black background are perceived as
hexahedrons. It seems to be well established that this
kind of illusion is totally caused by so-called
irradiation, i.e. the apparent expansion of light areas
(which can be given a simple, clear physical
explanation). "White circles expand due to irradiation
and reduce the black gaps between them", Professor
Paul Bert writes in his Lectures on Zoology. He goes on
to say that, "as each circle is surrounded by six other, it
pushes adjacent ones on expanding and appears to be
confined by a hexagon".
   Suffice it to glance at the neighbouring figure
(Fig. 141) where the same effect is observed for black
circles against a white background for this explanation
to be rejected: here irradiation only could reduce the
size of black spots but could not change them into
hexagons. For the two cases to be covered by the same
principle the following interpretation might be suggest-
ed. When viewing from a certain distance, the angle of
vision of the gaps between the circles becomes smaller
than a limit, so enabling their forms to be distinguish-
ed. Each of the six neighbouring gaps then appears to
be a straight line of a uniform thickness and the circles
are thus bounded by hexagons. This interpretation also
covers the paradoxical fact that at some distances white
circles continue to appear to be round, whereas the
black fringes around them have already assumed
hexagonal forms. It's only at larger distances that the
hexagonal configuration is transferred from the fringe
to the white spots. However, this explanation, too, is
only a plausible assumption and, perhaps, there are
several other possible explanations. It is necessary to
prove that the possible cause is here the actual one.
   Most of attempts to explain individual illusions
(except for the few mentioned above) are as unreliable
and uncertain. Some tricks of vision still await their
explanation. By contrast, others have too many
explanations each of which would perhaps be sufficient
in itself were there not so many additional ones that
make it less convincing. Remember the famous illusion
discussed since the time of Ptolemy-that of the
increasing of the size of celestial bodies near horizon.
 144-145     Optical   Illusions

             N o less than six possible theories, it seems, have been
             suggested, each of which has the only drawback that
             there are five more equally adequate explanations...
             Obviously, the entire domain of visual illusions is still
             in the pre-scientific stage of treatment and in need of
             establishing the basic methodology of its investigation.
                For want of any solid foundation in the form of
             relevant theories I have confined our discussion to the
             demonstration of unquestionable facts providing no
             explanations of what caused them and seeking only to
             present all the major types of visual illusion. * Only
             those involving portraits are explained at the end of the
             section since these are quite clear and incontestable...
             I also wanted to do away with some of superstitious
             notions that developed around this unique optical
                The series of illustrations opens with samples of
             illusions caused by clearly anatomical and physiological
             peculiarities of the eye. These are illusions due to the
             blind spot, irradiation, astigmatism, the retention of
             light impressions, and retina fatigue (see Figs. 100-107).
             In the blind spot experiment some of your field of
             vision may disappear in another way as well as
             Mariotte did for the first time in the 18th century. The
             effect perhaps is even more striking. So Mariotte writes:
             "Against a dark background approximately at the level
             of my eyes I attached a small circle of white paper and
             at the same time asked someone to hold another circle
             beside the first one about 2 feet to its right but
             somewhat lower so that the image would strike the
             optica] nerve of my right eye when I closed my left one.
             I stood next to the first circle and stepped back grad-
             ually without taking my right eye off it. When I was
             about 9 feet away, the second circle, which was about
             4 inches across, completely disappeared from my field of
                "I couldn't ascribe it to its lateral position, as I could
             discern other things further to the side than it. I'd have
             thought it removed had I not been able to find it again
             with the slightest movement of my eye...".
                These "physiological" tricks of vision are followed by
             a much larger class of illusions that are due to psycholo-
             gical reasons, which have not yet been sufficiently

                 * The selection of illusions here is the result of many years of
             collecting. I've excluded, however, all those published illusions that
             have effect not on anybody's eye or are not perceptible enough.
10 — 9 7 5
             Optical   Illusions

             clarified. It may perhaps be established that illusions of
             this kind are only the consequence of some
Figure 100   preconceived erroneous judgement that is involuntary
             and subconscious in nature. The source of the
             misperception here is the mind, not the sensor. Kant
             aptly remarked, "Our senses deceive us not because
             they do not judge correctly, but because they do not
             judge at all".

             Irradiation. When viewed from a distance the figures
             below-the circle and square-seem to be larger than
             those above, although they are equal in size. The larger
             the distance the more pronounced is the illusion. The
             phenomenon is called irradiation (see below).

             Irradiation. When viewed from a distance the figure
             with the black cross seems, owing to irradiation, to be
             distorted as shown in the accompanying figure on the
Figure 101
                Irradiation is due to the fact that each light point of
             an object produces on the retina of an eye not a point
             but a small circle because of so-called spherical
             aberration. Therefore a light surface on the retina is
             fringed by a light band that increases the place occupied
             by the surface. On the other hand, black surfaces
             produce reduced images because of the light band.

             The Mariotte Experiment. Close the right eye and look
             with the left one at the upper cross from a distance of
             20-25 centimetres. You'll notice that the middle, large
             white circle disappears completely, although the two
             smaller circles on either side are seen distinctly. If, with
             the same arrangement, you look at the lower cross, the
             circle only disappears in part.

Figure 102

               The phenomenon is caused by the fact that with this
             arrangement of the eye with respect to the figure the
             image of the circle falls on the so-called blind spot - t h e
             place insensitive to photic stimulation where the optic
             nerve is connected.
146-147      Optical   Illusions

Figure 103

             The Blind Spot. This experiment is a modification of
             the previous one. If you look at the cross at the right of
             the figure with your left eye at a certain distance you
             won't see the black circle at all, although the two cir-
Figure 104   cumferences will be seen.

             Astigmatism. Look at the lettering with one eye. Do all
             of the letters appear equally black? Normally one of
             the letters appears blacker than the rest of them. You
             need only to turn the page by 45° or 90° and some
             other letter will seem to be blacker.
               The phenomenon           is explained     by     so-called
             astigmatism, i.e. different curvatures of the retina in
             different directions (vertical, horizontal, etc.). It's only
             rarely that an eye is free of this imperfection.

Figure 105
             Optical   Illusions

             Astigmatism. Figure 105 furnishes another way (cf. the
             previous illusion) of identifying astigmatism in an eye.
             If you bring the figure to the eye under examination
             (the other one being closed) you'll notice at a certain,
             rather close, distance that two opposite sectors will
             seem blacker than the other two, which will appear
Figure 106

                When viewing this figure, move it to the right and
             left and it'll seem to you that the eyes in the figure
Figure 107
             swing horizontally.
                The illusion is accounted for by the eye's property to
             retain visual perceptions for a short time after the
             stimulus has disappeared (cinema is based on this).
                Having concentrated on the white square at the top
             you'll notice about half a minute later that the lower
             white line will have disappeared (owing to retina

             The Miiller-Lyer Illusion. The segment be seems to be
             longer than ab, although they are in fact equal.
             Figure 108

Figure 109
                       •<            >                      <
               Another form of the previous illusion is Fig. 109 and
             here segment A seems to be shorter than B.
               The deck of the right ship seems to be shorter than
             that of the left, actually both are the same length.

             Figure 110
148-149        Optical   Illusions

                 The distance AB seems to be much smaller than BC,
               which is equal to it.

Figure i l l

                             ^ > <
                  The distance AB seems to be larger than CD, which
Figure 112     is equal to it.

                                 c > — o

Figure 113                           • e e -
                 The lower oblong seems to be larger than the
               internal one, although these are equal (the influence of
               the arrangement).

                 The equal distances AB, CD and EF seem to be un-
               equal (the influence of the arrangement).

Figure 114

                  The rectangle crossed longitudinally seems to be
               longer and narrower than the equal rectangle crossed
               transversely (Fig. 115).
              Optical   Illusions

Figure 1 15

                 Figures A and B are equal squares, although the first
Figure 116    seems, to be higher and narrower than the second.

                The height of this figure seems to be larger than its
Figure 117
              width, although both are equal.

                 The height of the top hat seems to be longer than its
              width, although these are equal.
Figure 118       The distances AB and AC are equal, although the
              first seems to be longer.

              Figure 119

                 The distances BA and BC are equal, although the
              first seems to be longer.

Figure 120
                 Optical   Illusions
                   The upright plank seems to be longer than those
                 below, in fact these are all equal.
Figure 121

                   The distance MN seems to be smaller than the
                 distance AB, which is really equal to it.
                 Figure 122

Figure 123         The right circle in this figure seems to be smaller
                 than the equal-sized circle on the left.
                   The distance AB seems to be smaller than the equal
                 distance CD. This illusion becomes more pronounced
                 with distance (Fig. 124).

                   The empty gap between the lower circle and each of
                 the upper ones seems to be larger than the distance
                 between the outer edges of the upper circles. In actual
                 fact they are equal (Fig. 125).
Figure 124       Figure 125


                                          Figure 126

                 The "Smoking Pipe" Illusion. The dashes on the right
                 in this figure seem to be shorter than those on the left.
              Optical   Illusions

              The "Print Type" Illusion. The upper and lower parts
              of each of these characters seem to be equal to each
              other. But turn the page over and you'll immediately
              see that the upper parts are smaller.

              Figure 127

Figure 128

                The heights of the triangles are divided in two, but it
              seems that the part near the vertex is shorter.

              The Poggendorf Illusion. The oblique straight line
              intersecting the black and white strips seems to be
              broken from a distance.
              Figure 129

                If we continue both arches on the right, they will
              meet the ends of those on the left, although it seems
Figure I 31   that they should pass lower.

              Figure 130

                 Point c lying on the continuation of line ab, seems to
              lie below it.
152-153      Optical    Illusions

Figure 132     Both figures are identical, although the upper one
             seems to be the shorter and wider.
               The middle parts of those lines do not seem to be
             parallel, although they are so.

             Figure    133

             The Zollner Illusion. The long oblique lines of this
             figure are parallel, although these seem to be diverging.

             Figure 134

             The Hering Illusion. The two double parallel lines are
             parallel, although they seem to be arches with the
             crown facing each other (Fig. 135).
               The illusion disappears if (1) you hold the figure level
             to your eyes and view it so that you are glancing along
             the lines, or (2) you point the end of a pencil at some
             point and concentrate at this point.
             Optical   Illusions

Figure 135

               The lower arch seems to be more convex and shorter
             than the upper one. The arches are similar, though.

             Figure 136
Figure 137

               The sides of the triangle seem to be concave, in
             reality they are straight.
               The letters are upright type.
             Figure 138

               The curves in Fig. 139 seem to be a spiral, although
             they are circles, which is readily found by following
             anyone of them with a pencil.
             Optical   Illusions
Figure 139

Figure 140
             Optical   Illusions

               The curves in Fig. 140 seem to be oval, but
             they are circular, which can be tested with a pair of
               At a certain distance the circles in these figures
             (both white and black) seem to be hexagons.

Figure 141

             Autotype Illusion. Consider the pattern from a distance,
             and you will perceive an eye and part of the nose of
             a female face. The figure is a part of an autotype
             (conventional illustration in a book) multiplied tenfold.

Figure 142

                The upper silhouette seems to be longer than the
             lower one, although they are both the same size.
                Will the circle on the right of the figure get between
             lines AB and CDl It seems at first sight that it will. But
             really the circle is wider than the separation between
             the lines.
156-157      Optical   Illusions

Figure 143

Figure 144

               Distance AB seems to be wider than distance AC.
             which is equal to the former.

             Figure 145




                Holding Fig. 146 at eye level so that you glance
             along it, you'll see the picture given on the right.
                Close one eye and place the other approximately at
             the point where the continuations of these lines
             intersect. You'll see a row of pins as if stuck into the
              Optical   Illusions

Figure 146

             paper. Shift the figure slightly sidewards and the pins
             will swing.

             Figure 147                      Figure   148

Figure 149
                If you view this figure for a long time, it'll seem to
             you that the two cubes at the top and at the bottom
             stand out alternately. Also, you can intentionally, by
             exerting your imagination, call forth one or the other

             The Schroder Stairs. This figure might be perceived in
             three ways: (1) as stairs, (2) as a step-wise niche, or (3)
             as a pleated paper strip stretched out. The perceptions
             may interchange intentionally or unintentionally.
               The figure may represent, as you like it, either
             a block with a recess (the back side of the recess is the
             plane AB), or a part of an empty box with a block
158-159        Optica!    Illusions

Figure   150   touching the walls from the inside, the box being open
               at the bottom.
                  The intersections of the white lines in this figure seem
               to have yellowish square spots that appear and
               disappear, as if flashing. In actuality, the lines are
               absolutely white throughout, which can be seen if you
               cover adjacent rows of black squares with paper. The
               effect is because of the contrast.
               Figure 151

                 A modification of the illusion of Fig. 151, in which
               white spots appear at the intersections.
               Figure    152

                 Look at this figure from a distance. Its four strips
               each seem to be a concave stripe that is lighter at the
               edge and adjacent to a neighbouring, darker strip. But
               by masking neighbouring strips to exclude the influence
Optical   Illusions

of contrast you can see that each strip is uniformly
Figure 153

  Look attentively for a minute at some point on this
"negative" portrait of Newton without moving your
eyes, then quickly shift your glance to a piece of white
paper, greyish wall or ceiling. For a moment you'll see
the same portrait, but the black spots will become
white, and vice versa.

Figure 154

The Silvanus Thompson Illusion. If you rotate this
figure (by turning the book) all the rings and the white
toothed wheel will seem to be rotating, each about its
own centre, in the same direction and with the same
             Optical   Illusions

Figure 155

                On the left you see a convex cross, on the r i g h t - a
             concave one. But turn the book upside down and the
             figures will change their places. Actually the figures are
             identical, only they're shown at different angles.
Figure 156

                Look at the photograph in Fig. 157 with one eye
             14-16 centimetres away from the centre of the picture.
             With this arrangement your eye will see the picture
             from the same point the objective of the camera "saw"
             Optical   Illusions

Figure 157
             the scene. It's this that accounts for the liveliness of the
             impression. The landscape acquires depth, the water
                The eyes and the finger seem to point directly at you
             and follow you when you shift to the right or left.
                It has long been known that some portraits have the
             fascinating feature that they sort of follow the onlooker
             with their eyes and even turn their faces in his or her
             direction, wherever he or she shifts. This feature scares
             nervous people and is regarded by many as something
             supernatural. It has given rise to a number of
             superstitions and fantastic stories (e.g. The Portrait by
             Gogol). However, the explanation of this interesting
Figure 158   illusion is very simple.
                Above all, the illusion is peculiar not only to
             portraits, but to other pictures, too. A gun drawn or
             photographed so that it is directed at the onlooker*
             turns its muzzle in his direction when he moves to the
             right or left of the picture. Also, there is no evading
             a cart riding directly at the onlooker.
                All of these phenomena have one common and

                 * Such a photograph is obtained if, in photographing, the muzzle
             of the gun is directed at the objective. In exactly the same way if the
             person being photographed looks into the objective, then his eyes in
             the picture will be directed at the onlooker, at whatever angle he
             views the picture.
162-163   Optical   Illusions

          exceptionally simple cause. If we view the picture we
          imagine the things shown in it, and it seems to us that
          the thing has changed its position.
             The same applies to the portraits. When we observe
          a real face from the side, we see another part of it. We
          can only see the same part as before if the person turns
          his face to us, but in a portrait we always see the same
          view. When a portrait is perfectly executed the effect is
             Clearly, there is nothing surprising in this property of
          portraits. Conversely, it would be more unusual if, as
          we shift sidewards, we would see the side of the face.
          But, this, in essence, is what is expected by those who
          regard the apparent turn of the face in a portrait as
          something supernatural!
             Brain-Twisting Arrangements
             and Permutations

             In Six Rows
             You may have heard the funny story that nine horses
             have been put into 10 boxes, one in each. The problem
             that is now posed is formally similar to this famous
             joke, but it has a real solution *. You must arrange 24
             people in six rows with five in each.

             In Nine Squares
Figure 159   This is a trick question-half a problem, half a trick.
                Using matches make a square with nine small square
             cells and place a coin in each so that each row and
             column contain 6 kopecks (Fig. 159).
               The figure shows the arrangement of the coins. Place
             a match on one coin.
                Now ask your friends to change the arrangement
             without moving the coin with the match so that the
             rows and columns each still contain 6 kopecks.
                They'll say it's impossible. However, a small trick will
             help you to perform this "impossible" task. Which one?

             Coin   Exchange
             Make a large drawing of the arrangement in Fig. 160
             and denote each of the small squares by a letter in the
Figure 160   top left corner as shown. Put 1 kopeck, 2 kopeck, and
             3 kopeck coins into the three squares of the upper row.
             Now put 10 kopeck, 15 kopeck, and 20 kopeck coins
             into the three squares of the lower row. The rest of the
             squares are empty.
              By shifting the coins on vacant squares you make the
             coins exchange their places so that the 1 kopeck changes
             with the 10 kopecks, the 2 kopecks changes with
             the 15 kopecks, and the 3 kopecks with the 20 kopecks.
             You may occupy any vacant place of the figure but you
             are not permitted to place two coins into one square.
             Also, it isn't allowed to skip an occupied square or go
             beyond the figure.
              The problem is solved by a long series of moves.
             Which moves?

                 * In what follows the answers to problems are given at the end
             of each section.
164-165                          Brain-Twisting
                                 and Permutations

                                 Nine    Zeros
                                 Nine zeros are arranged as shown below:
                                 0 0 0
                                 0 0 0
                                 0 0 0
                                   You must cross all the zeros with four lines only.
                                   To simplify the solution I will add that the nine zeros
                                 are to be crossed without the pencil leaving the paper.

                                 Thirty Six Zeros
                                 You see that 36 zeros are arranged in the cells of this
      0      0   0   0   0   0
      0      0   0   0   0   0     You must cross out 12 zeros so that each row and
                                 column retain an equal number of uncrossed zeros.
      0      0   0   0   0   0     Which zeros are to be crossed?
      0      0   0   0   0   0
                                 Two     Draughtsmen
      0      0   0   0   0   0
                                 Put two different draughtsmen on a draughts board.
      0      0   0   0   0   0   How many different arrangements are possible?

                                 Flies on a Curtain
                                 Nine flies are sitting on a chequered window curtain.
                                 They happened to have arranged themselves so that no
                                 two flies are in the same row, column, or diagonal
                                 (Fig. 161).
Figure 161
             Brain-Twisting  Arrangements
             and Permutations

               After a while three flies shifted into neighbouring,
             unoccupied cells and the other six stayed in the same
             place. Curiously enough, the nine flies still continued to
             be arranged so that not a single pair appeared in the
             same direct or oblique line.
               Which three flies shifted and which cells did they

             Eight   Letters
Figure 162   The eight letters arranged in the cells of the square
             shown in Fig. 162 are to be arranged in alphabetical
             order by shifting them into a vacant cell, as in the two
             previous problems. This is not difficult if the number of
             moves is not limited, but you are required to achieve
             the result using a minimum number of moves. You
             must find out for yourself what the minimum number
             Squirrels and Rabbits
             Figure 163 shows eight numbered stumps. On stumps
             1 and 3 sit two rabbits, and on stumps 6 and 8, two
             squirrels. But both the squirrels and the rabbits are not
             happy with their seats and want to exchange them, the
             squirrels want to take the places of the rabbits, and the
             rabbits the places of the squirrels. They can only make
             it by leaping from a stump to the other along the lines
             indicated in the figure.

Figure 163
             Brain-Twisting  Arrangements
             and Permutations

               How could they make it?
               Observe the following rules:
               (1) each animal may make several leaps at once;
               (2) two animals may not seat on the same stump,
             therefore they must only leap on a vacant stump.
               Further, you should take into account that the
             animals want to reach their goal using the least
             possible number of leaps, although it's impossible to
             make less than 16 leaps.

             Cottage    Problem
Figure 164   The accompanying figure shows the plan of a small
             cottage whose poky rooms house the following furni-
             ture: a desk, a piano, a bed, a sideboard, and
             a bookcase. Only room 2 is free of furniture.
               The tenant wanted to change around the piano and
             the bookcase. This appeared to be a difficult problem
             because the rooms are so small that no two of the
             above pieces could be in the same room. The free room
             2 was of help. By shifting the things from one room to
             another the desired arrangement was eventually achiev-
               What is the least number of changes required to
             achieve the goal?

             Three     Paths
             Three brothers - Peter, Paul, and J a c o b - g o t three
             vegetable gardens located near their houses, as shown
             in the figure. You can see that the gardens are not very
             conveniently arranged but the brothers failed to agree
             about exchanging them.
                The shortest paths leading to the gardens crossed
             and the brothers began to quarrel. Wishing to avoid
             future conflict the brothers         decided     to find
             nonintersecting paths to their respective gardens. After
             a lot of searching they succeeded in finding such paths
             and now they come to their gardens without meeting
             each other.
                Could you indicate these paths?
                One requirement is that no path should go round
             Peter's house.
             Brain-Twisting  Arrangements
             and Permutations

Figure 165

                        Jacob's garden   Peter's garden   Paul s   garden

             Pranks of Guards
             The following is an ancient problem having many
             modifications. We'll discuss one of them.

Figure 166
168-165        Brain-Twisting
               and Permutations

                  The commander's tent is guarded by sentries housed
               in eight other tents (Fig. 166). Initially in each of the
               tents there were three sentries. Later the sentries were
               allowed to visit each other and their chief didn't punish
               them when, having come to a tent, he found more than
               three soldiers in it and less than three in the others. He
               only checked the total number of soldiers in each row
               of tents, thus if the total number of soldiers in the three
               tents of each row was nine, the chief thought that all of
               the guards were present.
                  Having noticed this the soldiers found a way to
               outwit their chief. One night four guards left and this
               passed unnoticed. On the next night six left and got
               away with that. On later night the guards began to
               invite guests: at one time four, at another eight, and at
               yet another, a full dozen guests. And all of these pranks
               passed unnoticed as the chief always found nine
               soldiers in the three tents of each row.
                  How did they manage to do so?

               Ten Castles
               In olden days a prince desired to have 10 castles built.
               They should be connected by walls arranged on five
               straight lines with four castles on each. The architect
               submitted the plan given in Fig. 167.
                  But the prince wasn't satisfied with the plan because
               the arrangement made all the castles vulnerable to
               outside attack, but he wished there to be at least one or

Figure   167
              Brain-Twisting  Arrangements
              and Permutations

              two castles protected within the walls. The architect
              objected that it was impossible to satisfy the condition
              whilst the 10 castles had to be arranged four in each of
              the five walls. But the prince insisted.
                 After a lot of head-scratching the architect in the
              long run came up with an answer.
                 Maybe you'll be happy enough, too, to arrange the
              10 castles and the frve interconnecting walls so as to
              meet the above conditions?

              An Orchard
              There were 49 trees in an orchard, arranged as shown
              in Fig. 168. The gardener decided that the orchard was
              too crowded, so he wanted to clear the garden of excess
              trees to make flowerbeds. He called in a workman and
              ordered: "Leave only five rows of trees, with four trees
              in each row. Cut down the rest and take them home for
              firewood as your payment for the work".
                 When the tree felling had finished the gardener came

Figure ! 68
170-165        Brain-Twisting
               and Permutations

               to see the result. Much to his dismay he found the
               orchard almost devastated: instead of the 20 trees the
               workman had left only 10 and cut 39.
                  "Why have you cut so many ? You were told to leave
               20 trees!" the gardener was enraged.
                  "No. You only told me to leave five rows with four
               trees in each. I did so. Just look."
                  The gardener was amazed to find that the 10
               remaining trees formed five rows with four trees in
               each. His order had been fulfilled literally, and still...
               39 trees had been cut down instead of 29.
                  How had the workman managed it?

               The White      Mouse
               All of the 13 mice in the figure are doomed to be eaten
               by the cat. But the cat wants to consume them in
               a certain order. The cat eats one mouse and then
               counts around the circle in the direction in which the
               mice are looking. When it gets to 13 it eats the mouse
               and starts counting again, missing out the eaten mice.
                Which mouse must it start from for the white mouse
               to be eaten last?

Figure   169

                               In Six Rows
The requirement of the problem is easily met if the people are arranged in the form of
a hexagon as shown in the figure.

Figure 170

                               In Nine Squares
You don't touch the forbidden coin but shift the whole of the lower row upwards
(Fig. 171). The arrangement has changed but the requirement of the problem is satis-
fied: the coin with the match hasn't been shifted.
Figure 171

                               Coin   Exchange
The following is the series of moves required (the number is the coin, the letter is the
cell to which the coin is shifted):
                                2-e   15—i     2-d    10-a
                               15—b    3-g     1 -h    3-e
                               10-d 20-c      10-e    15-b
                                2-h    1-e     2-j     2-d
                               20-e    3-a    15-i     3-j
                               10-j   15-b     3-g     2-i
  It's impossible to solve the problem in less than 24 moves.
172-173                        Answers

                              Nine       Zeros
The problem is solved as shown in Fig. 172.

Figure 172

                              Thirty Six         Zeros
As it's required to cross out 12 of the 36 zeros, we'll have 36—12, i.e. 24 zeros with
four zeros in each row.
  The remaining zeros will be arranged as follows:

                                0          0     0   0
                                           0     0   0   0
                                0    0     0             0
                                0    0           0       0
                                0    0               0   0
                                     0     0     0   0

                              Two     Draughtsmen
One draughtsman may be placed at any of the 64 squares of the board, i.e. in 64
ways, then the second one can occupy any of the 63 remaining squares. Hence for
each of the 64 positions of the first draughtsman you can find 63 positions for the
second one. Consequently, the total number of the various permutations of the two
draughtsmen is:

                                     64 x 63 = 4,032.

                             Flies on a Curtain
The arrows in Fig. 173 indicate which flies must be shifted and in which direction.

Figure 173
                             ' / i l i i i i M

                             Eight     Letters
The least number of moves is 23. These are as follows:
             A B F E C A B F E C A B D H G A B D H G D E F

                             Squirrels and Rabbits
Shown below is the shortest way of the rearrangement The first number in each pair
indicates from which stump an animal should leap and the second number the
destination stump (for example, 1-5 means that a squirrel has leapt from the first
stump to the fifth). The total number of leaps required is 16, namely:
                      1-5; 3-7; 7-1; 5-6; 3-7; 6-2; 8-4; 7-1;
                      8-4; 4-3; 6 - 2 ; 2-8; 1-5; 5-6; 2-8; 4-3.

                            Cottage Problem
The exchange can be achieved in no less than 17 moves. The pieces of furniture are
moved in the following sequence:
            1. Piano               7. Piano           13. Bed
            2. Bookcase            8. Sideboard       14. Sideboard
            3. Sideboard           9. Bookcase        15. Table
            4. Piano              10. Table           16. Bookcase
            5. Table              11. Sideboard       17. Piano
            6. Bed                12. Piano

                             Three Paths
The three nonintersecting paths are shown in Fig. 174.
                                                         Peter's house

Figure   174

                                 Jacob's garden        Peter's garden           Paul's garden

                                Pranks of Guards
The problem is easily solved by the following reasoning. For four guards to be able to
be absent unnoticed by the chief it's necessary that in rows / and III (Fig. 175«) there
are nine soldiers in each. As the total number is 24 — 4 = 20, then in row II there will

Figure   175                          IV   V      VI

                                 ' • • •                      • • •                             • • •
                                "•••                         0 B0
                                                             0 0 0 000      b                        c

                                  0        0 0               0 0 0                        •     Q        •
                                  •        • 0               0 * 0                              0 « 0
                                  000      d
                                                            000         e
                                                                                            O00  f

be 20 - 18 = 2, i.e. one soldier in the left tent of this row and one in the right. In the
same way we find that there must be one soldier in the upper tent of row V and one
in the lower. It is now clear that the corner tents must house four guards. Accordingly,
the required arrangement for four soldiers to be absent is as shown in Fig. 175b.
   A similar argument yields the desired arrangement for six soldiers to be absent
(Fig. 175c).
   For four guests the arrangement is shown in Fig. 175d;
   For eight guests in Fig. 175e;
   And finally, Fig. 175/ shows the arrangement for 12 guests.
   It is easy to see that under these conditions no more than six soldiers can be absent
with impunity and no more than 12 guests can visit the guards.

                                Ten Castles
Figure 176 (on the left) shows the arrangement with two castles protected from the
external attack. You see that the 10 castles are disposed as required in the problem:

Figure 176

four on each of the five lines. On the right of Fig. 176, four more solutions to the
problem are given.

                              An Orchard
The uncut trees were disposed as given in Fig. 177. These form five straight rows with
four trees in each.
Figure   ill
176-177                        Answers

                               The White   Mouse
The cat should first eat the mouse at which it is looking, i. e. the sixth one from the
white. Try it by beginning with this mouse and cross out every 13th mouse. You'll see
that the white mouse will be the last to be crossed out.
                 Skilful Cutting and Connecting

             7   With Three Straight   Lines
                 Cut Fig. 178 into seven sections with three straight
Figure 178       lines so that there is one animal in each section.

Figure 179

                 Into Four Parts
                 This ground area (Fig. 179) consists of five equal
                 squares. Draw the area on a sheet of paper. Can you
                 cut it into four identical areas, not five?

                 To Make a Circle
                 A joiner was given two pieces of rare wood with holes
                 in them (as shown) and was asked to make them into
Figure 180       a perfectly circular solid board for a table so that no
                 scraps of the expensive wood would be left over. All the
                 wood must be used.
                    The joiner was a master craftsman but the order was
                 not easy. He scratched his head for a long time, tried
                 one way and then another, and eventually hit upon an
                 idea as to how to execute his order.
                    Perhaps you'll twig it, too? Cut out two paper
                 figures, exactly like the ones in Fig. 180 (only larger)
                 and use them to arrive at the solution.

                 A Clock Dial
                 The clock dial in Fig. 181 must be cut into six parts of
                 any shape so that the sum of numbers in each section
178-179        Skilful Cutting and   Connecting

Figure 181
               would be the same.
                 The aim of the problem is not so much to test your
               resourcefulness but the quickness of your thought.

               The crescent (Fig. 182) must be divided into six parts
               by only two straight lines.

Figure   182   To Divide a Comma
               In the accompanying figure you will see a wide comma.
               It's constructed very simply: a semicircle is drawn on
               the straight line AB around point C, then two
               semicircles are drawn around the middles of the
               segments AC and CB, one on the right, the other on
               the left.
                  You must cut the figure into two identical parts by
               a single curved line.
Figure 183
                  The figure is also interesting in that two such figures
               make up a circle. How?

               To Develop a Cube
               If you cut a cardboard cube along edges so that it
               could be unfolded and placed with all six squares on
               a table, you'll get a figure like one of those shown in
               Fig. 184.
               Figure   184

                 It's curious, but how many different figures can be
               obtained in this way? In other words, in how many
               ways can a cube be developed?
                 I warn the impatient reader that there are no less
               than 10 different ways.

               To Make Up a Square
               Can you make up a square from five pieces of paper
               like the ones shown in Fig. 185a?
                  If you've already found the solution, try and make up
             Skilful Cutting   and   Connecting

             a square from five identical triangles like the ones you
             have just used (the base is twice as long as the height).
             You may cut one of the triangles into two parts but the
             other four must be used as they are (Fig. 185b).
Figure 185
                              With Three Straight Lines

The problem is solved as follows:

Figure 186

                              Into Four Parts
The dash lines show the way in which the ground must be divided (Fig. 187).

Figure 187

                               To Make a Circle
The joiner has cut each of the boards into four parts as shown on the left of Fig. 188.
From the four smaller parts he makes up a smaller inner circle to which he glues the
other four parts. He thus got an excellent board for a round table.

Figure 188

                               A Clock Dial
As the sum of all the numbers on the face of the dial is 78, the sum of each of the six
sections mast be 78-f-6 = 13. This facilitates finding the solution that is shown in
Fig. 189.

Figure 189                        J ^ T j Z ^ K .

The answer is shown in the accompanying figure. The resultant six parts are
Figure 190

                               To Divide a Comma
The solution is seen in the accompanying drawing. Both parts of the comma are
equal, as they are made up of equal parts. The figure shows how the circle is made of
two commas, one white and one black.
Figure 191

                              To Develop a Cube
All the 10 possible solutions are shown in Fig. 192. The first and fifth figures can be
turned upside down and this will add two more involutes, increasing the total to 12.
Figure 192

                              To Make Up a Square
The solution of the first problem is shown in Fig. 193a. The case of triangles is given
in Fig. 193b. One triangle is first cut up as shown.

Figure 193
             Problems with Squares

             A Pond
             There is a square pond (Fig. 194) with four old oaks
             growing at its corners. It is required to expand the

Figure 194

             pond so that its surface area be doubled, the square
             shape being retained and the old oaks not destroyed or

             A Parquet   Maker
             When cutting wooden squares a parquet maker tested
             them thus: he compared the lengths of sides and if all
             four sides were equal he considered the square to be
             cut correctly.
               Is this test reliable?

             Another Parquet     Maker
             Another parquet maker checked his work otherwise: he
             measured diagonals not the sides. If both diagonals
             were equal, he considered the square to be true.
Figure 195     Are you of the same opinion?

             Yet Another Parquet    Maker
             Yet another worker checked his squares by seeing if all
             the four sections into which the diagonals divide each
             other (Fig. 195) are equal to each other. In his opinion
             it proved that the rectangle cut was square.
                What do you make of that?
             Problems with Squares
             A Seamstress
             A seamstress wants to cut out a piece of linen in the
             form of a square. Having cut several pieces she checks
             her work by bending each piece along its diagonal to
             see if the edges coincide. If they do, she thinks, each
             piece is perfectly square.
               Is she right?

             Another   Seamstress
             Another seamstress wasn't satisfied with the check her
             companion used. She bent her piece first along one
             diagonal and then after smoothing the linen she bent it
             along the other. It was only if the edges of the piece
             coincided in both cases that she thought the square was

               What would you say about this test?

             A Joiner's Problem
Figure 196
             A young joiner has the five-sided board shown in
             Fig. 196. You see that it seems to be composed of
             a square glued to a triangle that is four times smaller
             than the square. The joiner is asked to make the board
             into a square, taking nothing away from the board and
             adding nothing to it. This, of course, involves cutting it
             into sections. O u r young joiner is just going to do so,
             but he wants to cut the board along no more than two
             straight lines.
                Is it possible, using two lines, to cut the figure into
             parts from which the joiner could make a square? And
             if the answer is "yes" how does he go about it?
                    n          Answers

                               A Pond
It is possible to double the surface area of the pond with the square shape retained
and the oaks intact. The accompanying figure shows how this can be done. You can

Figure 197

easily see that the new area is twice the earlier, just draw in the diagonals of the
earlier pond and count the resultant triangles.

                               A Parquet    Maker
The test is not sufficient. Some quadrilaterals that are by no means squares will pass.
Figure 198 gives examples of quadrilaterals whose sides are equal but whose angles
are not right (rhombs).

Figure 198

                               Another Parquet      Maker
This test is as unreliable as the first one. To be sure, a square's diagonals are equal
but not every quadrilateral with equal diagonals is a square. It is clearly seen from the
examples in Fig. 199.
Figure 199

  The parquet makers should apply both tests to each quadrilateral produced. One
could then be sure that the work has been done correctly. Any rhomb with equal
diagonals is bound to be a square.
186-187                           Answers

                                  Yet Another Parquet      Maker
The test might only show that the quadrilateral in question has right angles, i. e. that
it is a rectangle. But it fails to verify that all its sides are equal, as is seen in Fig. 200.

Figure 200

                                 A Seamstress
The test is far from adequate. Figure 201 presents several quadrilaterals whose edges
coincide when bent along the diagonals, yet they are not squares. You see how far
Figure 201

a quadrilateral may differ from a square and still satisfy this test.
  The test only shows that the figure is symmetrical, no more.

                                 Another    Seamstress
This test is no better than the previous one. You could cut any number of
quadrilaterals out of paper that would pass this test, although they are by no means
squares. The examples in Fig. 202 all have equal sides (these are rhombs) but the
Figure 202

angles are not right-hence these are not squares.
  In order to make really sure that the pieces cut out are squares, the seamstress
should additionally check if the diagonals (or angles) were equal.

                                  A Joiner's   Problem
One line should go from the vertex c to the middle of side de, the other, from the last
point to vertex a. A square can be made up from the three pieces 1, 2, and 3 as shown
in Figure 203.                      c
Figure 203
             Problems on Manual Work

             Five navvies excavate a 5-metre ditch in 5 hours. How
             many navvies are required to dig 100 metres of ditch in
             100 hours?

             A lumberjack cuts a 5-metre log into 1-metre lengths. If
             each cut takes 1.5 minutes, how long will it take to cut
             the log?

             Joiner and Carpenters
             A team of six carpenters and a joiner did a job. Each
             carpenter earned 20 roubles, but the joiner got
             3 roubles more than the average earnings of all the
             seven team members.
               How much did the joiner earn?

             Five Pieces of Chain
Figure 204   A blacksmith was given five pieces of chain with      three
             links in each (Fig. 204) and asked to connect        them.
                The blacksmith opened and reclosed four           links.
                But is it not possible to do the same job with    fewer
             links tampered with?

             How Many      Vehicles?
             A shop repaired 40 vehicles (cars and motocycles) in
             a month. The total number of wheels on the vehicles
             was 100.

               How many cars and motocycles were              repaired?

             Potato Peeling
             Two people peeled 400 potatoes. One completed three
             pieces a minute, the other two. The second worked 25
             minutes longer than the first.
                How long did each work?
             Two   Workers
             Two workers can perform a job in seven days provided
             the second starts two days later than the first. If the job
188-189   Problems on Manual   Work

          were done by each of them separately, then the first
          would take four days more than the second.
            How many days would each of them take to perform
          the job individually?
            The problem permits of a purely arithmetic solution
          without any need to manipulate fractions.

          Typing a Report
          Two typists type a report. The more experienced one
          could finish the work in 2 hours, the other in 3 hours.
            How long will it take them to do the job if they
          divide it so as to spend the least time possible?
            Problems of this kind are normally solved according
          to the procedure of the famous problem on reservoirs.
          Thus in our problem they would find the share of the
          work done by each typist, add up the fractions and
          divide unity by the resultant sum.
            Could you think of some other procedure?

          Weighing Flour
          A salesman has to weigh five bags of flour. His problem
          was that the shop had a balance but some weights were
          missing so that it was impossible to weigh from 50 to
          100 kilogrammes. But the bags weighed 50-60
          kilogrammes each.
             The man began to weigh the bags in pairs. Of the
          five bags it is possible to make 10 different pairs, so he
          had to make 10 weighings. He produced the series of
          numbers given below in the ascending order:
          110 kg,    112 kg,    113 kg,   114 kg,   115 kg,
          116 kg,    117 kg,    118 kg,   120 kg,   121kg.
            How much did each bag weigh?

                        vJ'     Navvies
It's easy to swallow the bait and think that if five navvies dug 5 metres of the ditch in
5 hours, then it would take 100 people to dig 100 metres in 100 hours. But that
argument is absolutely wrong, since the same five navvies would be required, no more.
   In fact, five navvies dig 5 metres in 5 hours, so they can do 1 metre in 1 hour, and
in 100 hours-100 metres.

The common answer would be 1.5 x 5, i.e. 7.5 minutes. That is because many people
often forget that the last cut will give two 1-metre lengths. Thus, it's only necessary to
cut the log four times, not five, and this will take 1.5 x 4 = 6 minutes.

                               Joiner and Carpenters
We can easily find the average earnings of a member of the team by dividing the extra
3 roubles between the six carpenters. Accordingly, we should add 50 koppecks * to the
20 roubles earned by each carpenter to arrive at the average earnings of each of the
seven workers.
  We'll thus obtain that the joiner earned 20 roubles 50 kopecks plus 3 roubles, i.e.
23 roubles 50 kopecks.

                               Five Pieces of Chain
It's only necessary to open the three links of one of the pieces and to use the links
obtained to connect the other four pieces.

                               How Many      Vehicles?
If all the 40 vehicles were motocycles, the total number of wheels would be 80, i. e. by
20 less than in reality. Replacing a single motocycle by a car increases the total
number of wheels by two and the difference decreases by two. Clearly, 10 such
replacements are required for the difference to be reduced to zero. So, there were 10
cars and 30 motocycles.
   In fact: 1 0 x 4 + 3 0 x 2 = 100.

                               Potato Peeling
During the 25 extra minutes the second peeler put out 2 x 25 = 50 pieces. We subtract
50 from 400 to find that if the two had worked an equal time they would have yielded
350 potatoes. As their production per minute was 2 + 3 = 5 pieces, then by dividing
350 by 5 we find that each would have worked for 70 minutes.

  * 1 rouble = 100 kopecks.
  This is the actual duration of work of the first peeler, the second one worked for
70 + 25 = 95 minutes.
  In fact: 3 x 70 + 2 x 95 = 400.

                                Two       Workers
If each worker performs half the job individually, the first would need two days more
than the second (because the difference in duration for the whole job is four days). As
in our case the difference is just two days when the two work together, it is then
obvious that during the seven-day period the first worker performs half the job,
whereas the second does his half in five days. Thus, the first worker would be able to
do the whole job himself in 14 days and the second in 10 days.

                                Typing a Report
A nonstereotyped approach is as follows. First, we'll ask the question: if the typists are
to finish the work simultaneously, how should they divide it? (Clearly, it's only under
this condition, i. e. without any time wasted, that the work will be done in the shortest
time possible). As the more experienced typist types 1.5 times faster it's obvious that
her share should be 1.5 times larger than that of the other if both are to stop
simultaneously. It follows that the first typist should take over three fifths of the
report, and accordingly the second two fifths.
   As a matter of fact the problem is nearly solved. It only remains to find the time
taken by the first typist to do her share of the job. We know she can do the whole job
in 2 hours, hence the three fifths of the job will be carried out in 2 x 3/5 = 11/5 hours.
During exactly this time the second typist will finish her share of the job.
   Thus, the shortest time required for both typists to type the report is 1 hour and 12

                                Weighing Flour
To begin with, the salesman summed up the 10 numbers. The resultant sum (1,156
kilogrammes) is nothing but the fourfold weight of the bags: the weight of each bag
enters the sum four times. If we divide ,by four, we'll find that the total weight of the
five bags is 289 kilogrammes.
   We'll now for convenience assign numbers to the bags in ascending order of their
weights. The lightest bag will be No. 1, the second No. 2, etc., and the heaviest,
No. 5. It will be seen that in the series of quantities: 110 kg, 112 kg, 113 kg, 114 kg,
115 kg, 116 kg, 117 kg, 118 kg, 120 kg, and 121 kg, the first quantity is the sum of the
weights of the two lightest bags, No. 1 and No. 2; the second quantity, of No. 1 and
No. 3. The last quantity (121) is the sum of the two heaviest bags, No. 4 and No. 5,
and the penultimate, of No. 3 and No. 5. Thus:
                                No.   1   and   No.   2 give 110 kg
                                No.   1   and   No.   3 » 112 kg
                                No.   3   and   No.   5 » 120 kg
                                No.   4   and   No.   5 » 121 kg

   We can thus easily find the sum of the weights of No. 1, No. 2, No. 4, and No. 5:
110 kg + 121 kg = 231 kg. Subtracting this number from the total weight of the bags
(289 kg) gives the weight of No. 3, namely-58 kg.
   Further, from the sum of No. 1 and No. 3, i.e. from 112 kg, we subtract the
now-known weight of No. 3 to arrive at the weight of No. 1: 112 — 58 = 54 kg.
   In exactly the same way we find the weight of No. 2 by subtracting 54 kg from
 110 kg, i.e. from the sum of No. 1 and No. 2. The weight of No. 2 will thus be 110 —
 - 54 = 56 kg.
   Now from 120 kg (No. 3 + No. 5) we subtract the weight of No. 3 (58 kg) to get
the weight of No. 5: 120 - 58 = 62 kg.
   It remains to determine the weight of No. 4, knowing the sum of No. 4 and No. 5
(121 kg). Subtracting 62 from 121 gives that No. 4 weighs 59 kg.
   The weights of the bags are thus
   54 kg, 56 kg, 58 kg, 59 kg, 62 kg.
   We have solved the problem without any resort to equations.
192-193   Problems on Purchases
          and Prices

          How Much are the Lemons?
          Three dozen lemons cost as many roubles as one can
          have lemons for 16 roubles.
            How much does a dozen lemons cost?

          Raincoat, Hat and Overshoes
          A raincoat, hat and overshoes are bought for 140
          roubles. The raincoat costs 90 roubles more than the
          hat, and the hat and the raincoat together cost 120
          roubles more than the overshoes.
            How much does each thing cost separately?
            Use mental arithmetic only, no equations.

          When I went out shopping I had in my purse 15
          roubles in 1 rouble pieces and 20 kopeck coins. When
          back home I had as many 1 rouble pieces as there had
          been 20 kopeck coins initially, and as many 20 kopeck
          coins as I had had 1 rouble pieces initially, my purse
          only containing a third of the initial sum
            How much had I spent?

          Buying Fruit
          One hundred pieces of various fruit can be bought for
          five roubles. The prices are: water-melons, 50 kopecks
          a piece; apples, 10 kopecks a piece; and plums, 10
          kopecks a ten.
             How many fruit of each kind are bought?

          Prices Up and Down
          The price of a product first went up 10%, and then
          down 10%.
            When was the price lower, initially or finally?

          Six barrels of beer were shipped to a shop. The
          numbers in Fig. 205 show the numbers of litres in each
          barrel. Two customers bought five of the six barrels,
             Problems on Purchases
             and Prices

Figure 205

             one bought two and the other bought three. Given that
             the second bought twice as much beer as the first,
             which barrel wasn't sold?

             Selling Eggs
             At first sight, this ancient problem might seem incon-
             gruous as it involves selling half an egg. Nevertheless, it's
             quite solvable.
               A peasant woman came to a market to sell some
             eggs. A first buyer took half her eggs plus 1/2 of an egg.
             A second buyer bought half the remaining eggs plus
             another 1/2 of an egg. A third only bought one egg,
             which was the last.
               How many eggs were there initially?

             Benediktov's   Problem
             Many experts in Russian literature don't suspect that
             the poet V. G. Benediktov (1807-1873) was also the
             author of the first collection of mathematical
             brain-twisters in the language. The collection wasn't
             printed and remained in a manuscript form to be found
             only in 1924. I had the opportunity to get acquainted
             with the manuscript and even established, based on one
             of the problems, the year it was compiled, namely 1869
             (the manuscript wasn't dated). The problem given
             below has been treated by the poet and named "An
             Ingeneous Solution of a Difficult Problem".
                "An egg seller sent her three daughters to the market
             with ninety eggs. She gave ten to the eldest and
194-187   Problems on Purchases
          and Prices

          cleverest daughter, thirty to the second, and fifty to the
          third, saying:
             'You should agree beforehand on the price at which
          you'll sell the eggs and stick to it. All of you should
          adhere to this price but I hope that the eldest daughter
          who is so bright will nevertheless be able to get as
          much for her ten eggs as the second daughter will
          receive for her thirty and she will teach the second
          sister how to get as much for her thirty as the youngest
          sister will get for her fifty eggs. Let the takings and
          prices be the same for the three of you. Furthermore,
          I'd like you to sell the eggs so that on average you will
          receive no less than 10 kopecks for ten, and no less
          than 90 kopecks for the ninety!'"
             Here I interrupt Benediktov's story so that the
          readers could figure it out for themselves how the girls
          went about their business.

                   1Q ^         How Much are the Lemons?
We know that the 36 lemons cost as many roubles as they sell lemons for 16 roubles.
But 36 lemons cost
                            36 x (price of one lemon).
For 16 roubles one can have
                                16/(price of one lemon).
                                36 x (price of one lemon) = 16/(price of one lemon).
After some algebra we have
                                (price of one lemon) x (price of one lemon) = 16/36.
   Clearly, one lemon costs 4/6 = 2/3 rouble and a dozen lemons cost 2/3 x 12 = 8

                                Raincoat, Hat and Overshoes
If instead of the raincoat, hat, and overshoes only two pairs of overshoes were bought,
the price would be not 140 roubles, but 120 roubles less. Thus, the two pairs of
overshoes cost 140— 120 = 20 roubles, hence one pair cost 10 roubles.
   Now we find that the raincoat and the hat together cost 140 — 10 = 130 roubles, the
raincoat costing 90 roubles more than the hat. We argue as earlier: instead of the
raincoat and hat we could buy two hats, and we would pay not 130 roubles but 90
roubles less, i. e. 130 — 90 = 40 roubles. Hence one hat costs 20 roubles.
   Thus, the prices of the things were as follows: the overshoes-10 roubles, the h a t - 2 0
roubles, and the raincoat-110 roubles.

Denote the initial number of 1 rouble pieces by x, and the number of 20 kopeck coins
by y. Then when I went out shopping I had in my purse
                               (lOOx + 20y) kopecks.
  Back from my shopping expedition I had
                          (lOOy + 20x) kopecks.
  As stated, the    latter   sum is three times smaller than         the   former,   hence
                                3 (100y + 20x) = lOOx 4- 20y.
  Rearranging the expression gives
                               x = ly.
  If y = 1, then x = 7. Under this assumption I initially had 7 roubles 20 kopecks
which is at variance with the statement of the problem ("about 15 roubles").
   Let's try y = 2, this gives x = 14. The initial sum is thus 14 roubles 40 kopecks
which checks well with the problem statement.
  The assumption of y = 3 leads to an overestimation: 21 roubles 60 kopecks.
   In consequence, the only fitting answer is 14 roubles 40 kopecks. When I returned
back from my shopping excursion I only had two 1 rouble pieces and fourteen 20
kopeck coin, i. e. 200 + 280 = 480 kopecks, which actually amounts to a third of the
initial sum (1,440/3 = 480).
   As I spent 1,440 — 480 = 960 kopecks, my purchases had cost 9 roubles 60 kopecks.

                               Buying Fruit
Despite the seeming uncertainty the problem has the only solution:
                                 Number             Cost
                Water melons      1                           50 kopecks
                Apples           39                 3 roubles 90 kopecks
                Plums            60                           60 kopecks

                Total           100                 5 roubles 00 kopecks

                                Prices Up and Down
It would be erroneous to consider that the two prices are equal. It's easily shown that
this is not the case. After the price went up the article cost 110%, or 1.1 of the initial
price. But after the price went down it amounted to
                                1.1 x 0.9 = 0.99,
i.e. 99% of the initial price. Consequently, the final price was 1% lower than the initial

The first customer bought the 15 litre and 18 litre barrels and the second -the 16 litre,
19 litre and 31 litre barrels. Really,
                                      15 + 18 = 33
                                1 6 + 19 + 31 = 66,
i. e. the second customer bought twice as much beer as the first one. The 20 litre barrel
remained unsold.
    This is the only possible solution as no other combination gives the relationship

                               Selling Eggs
The problem is worked out backwards from the end. After the second buyer bought
half the remaining eggs plus 1/2 of an egg, there was only one egg that remained with
the peasant woman. Accordingly, 1 1/2 eggs was half of what remained after the first

sale and so the full number is three eggs. We add 1/2 of an egg to obtain half of what
the woman had initially. Thus, the woman had brought seven eggs for sale.
   Let's check:
                                7/2 = 3 1/2; 3 1/2 + 1/2 = 4; 7 - 4 = 3
                                3/2 = 1 1/2; 1 1/2 + 1/2 = 2; 3 - 2 = 1,
which complies with the conditions of the problem.

                                Benediktov's   Problem
We continue the interrupted story:
   "The problem was a very difficult one. The sisters put their heads together on their
way to the market, the two younger sisters seeking advice of the eldest. The latter gave
some thought to the matter and said:
   'Sisters, we'll sell the eggs not by the ten, as is the custom here, but by the seven.
And we'll set a price for the seven we'll stick to as Mother said. Not a kopeck down
from the set price! The first seven goes for three kopecks, agreed?
   'Dirt-cheap', the second sister said.
   'But', the eldest sister continued, 'we'll raise the price for those eggs that'll remain
after we have sold the full sevens. I've checked beforehand that there'll be no other egg
sellers in the market. No one to beat down the price. But when there is demand and
the supply is dwindling the price rises. So we'll make up for our loss with the
remaining eggs'.
   'And what will we charge for the remaining eggs?5 the youngest sister asked.
   'Nine kopecks for each egg. Cash down! Those who need eggs badly will pay'.
   'Rather dear,' the second sister noted again.
   'What of i t ? the eldest said, 'the first eggs will have been sold cheaply by the seven.
One will compensate for the other!'
   "Understandably, the first to go were the fifty eggs of the youngest sister. She
received 21 kopecks for 7 sevens and one egg remained in her basket. The second one
sold 4 sevens for 12 kopecks and two eggs remained in her basket. The eldest sister
sold a seven for 3 kopecks and three eggs remained in her basket.
   "The last six eggs were sold for nine kopecks each. So the eldest got 27 kopecks for
her three eggs which brought her takings to 30 kopecks. The second sister got 18
kopecks for her last pair of eggs which when added to the 12 kopecks received earlier
for her 4 sevens, gave her 30 kopecks as well. The youngest sister got 9 kopecks for
her single egg and when she added the money to the 21 kopecks for her 7 sevens her
total was 30 kopecks, too.
   "Thus, the money they got for ten appeared to be equal to the money they got for
             Weight and Weighing
             One Million Times the Same Product
             A product weighs 89.4 grammes. Figure out how many
             tonnes a million of them weigh.

             Honey and Kerosene
             A jar of honey weighs 500 grammes. The same jar filled
             with kerosene weighs 350 grammes. Honey is twice as
             heavy as kerosene.
               What is the weight of the empty jar?

             A Log
             A round log weighs 30 kilogrammes.
               How much would it weigh if it were twice as thick,
             but twice as short?

             Under Water
             Consider a balance on the one pan of which there is
             a boulder that weighs exactly 2 kilogrammes and on
             the other, an iron weight of 2 kilogrammes. I carefully
             immerse the balance in water.
               Will the pans be in equilibrium?

             A Decimal Balance
             A decimal balance weighs 100 kilogrammes of iron
             nails that are balanced by iron weights.
               When submerged, will the balance be in equilibrium?
Figure 206
             A Piece of Soap
             Onto one pan of a balance a piece of soap was put,
             onto the other 3/4 of a same sized piece plus 3/4
             kilogramme. The balance is in equilibrium.
               What is the weight of a whole piece?
                Try and solve the problem mentally, without a pencil
             and paper.

             Cats and   Kittens
             The accompanying figure shows that the four cats and
             three kittens together weigh 15 kilogrammes and that
                          Weight and   Weighing

             Figure 207

                          three cats and four kittens weigh 13 kilogrammes. All
Figure 208                the cats have the same weight, so do the kittens.
                            How much does one cat weigh? And a kitten?
                             This problem, too, should be solved mentally.

                          Shell and Beads
                          Figure 208 shows that three children's blocks and one
                          shell are balanced by 12 beads and further that one
                          shell is balanced by one block and eight beads.
                            How many beads should be placed on the vacant
                          pan for the shell on the other pan to be balanced?

                          A further problem of the same kind. It is seen in Fig.
                          209 that three apples and one pear weigh as much as
                          10 peaches, but six peaches and one apple weigh as
Figure 209                much as one pear.
                            How many peaches are required to balance one

                          How Many       Glasses?
                          You see in Fig. 210 that a bottle and a glass are
                          balanced by a jug, the bottle is balanced by a glass and
                          a saucer, and two jugs are balanced by three saucers.
                            How many glasses should be placed on the vacant
                          pan for the bottle to be balanced?

                          With a Weight and a Hammer
                          It's required to weigh out 2 kilogrammes of sugar into
                          200-gramme packets. There is, however, only one
             Weight and   Weighing
Figure 210   Figure 211

             500-gramme weight and hammer that weighs 900
               How should one go about it using the weight and the

             Archimedes's     Problem
             The most ancient of brain-twisters pertaining to
             weighing is undoubtedly the one the tyrant of Syracuse
             Hieron gave to the famous mathematician Archimedes.
                The legend has it that Hieron entrusted a craftsman
             to manufacture a crown for a statue and ordered to
             give him the required amount of gold and silver. When
             it was ready, the crown weighed as much as the initial
             amounts of gold and silver had originally weighed
             together, but the craftsman was alleged to have stolen
             some of the gold having replaced it by silver.

               Hieron called in Archimedes and asked him to
             determine how much gold and silver respectively the.
             crown contained.
               Archimedes solved the problem proceeding from the
             fact that in water pure gold loses one twentieth of its
             weight, and silver one tenth.
                If you want to try your hand at the problem suppose
             that the craftsman was given 8 kilogrammes of gold
             and 2 kilogrammes of silver and when Archimedes
             weighed the crown under water the result was 9 1/4
             kilogrammes, not 10 kilogrammes. Given that the
             crown was made of solid metal, without any voids, how
             much gold had the craftsman stolen?

                                One Million Times the Same Product
The mental arithmetic here is as follows. We must multiply 89.4 grammes by one
million, i.e. by one thousand thousands.
  We can do the multiplication in two steps: 89.4 x 1,000 = 89.4 kilogrammes because
the kilogramme is 1,000 times larger than the gramme. Then, 89.4 kilogrammes x
 x 1,000 = 89.4 tonnes, because the tonne is 1,000 times larger than the kilogramme.
  The weight we seek is thus 89.4 tonnes.

                                Honey and Kerosene
Since honey is twice as heavy as kerosene, the difference in weight (500 — 3 5 0 = 150
grammes) is the weight of the kerosene in the volume of the jar (the jar of honey
weighs as much as a jar containing a double aijiount of kerosene). Hence we
determine the weight of the jar: 3 5 0 - 150 = 200 grammes. Really: 500 - 200 = 300
grammes, i.e. the honey is two times heavier than the same amount of kerosene.

                                A Log
A common answer is that a log, whose thickness has increased twice, and length
decreased twice, should be the same weight. This is not so, however. Doubling the
diameter increases the volume of a round log fourfold, but halving its length halves its
volume. The net result is that the final log is twice as heavy as the initial one, i. e. it
weighs 60 kilogrammes.

                                Under Water
Each immersed body becomes lighter by the weight of the water displaced by it. This
law, discovered by Archimedes, will help us to answer the problem.
   The 2-kg boulder has a larger volume than the 2-kg iron weight because the
material of the boulder (granite) is lighter than iron. Accordingly, the boulder will
displace a larger volume of water than the weight and, according to Archimedes's
principle, loses more than the weight. The weight will thus outweigh the boulder
under water.

                                Decimal Balance
When immersed in water, an iron object loses one eighth of its weight.* Thus both the
nails and the weights will when immersed have only 7/8 of their former weight. Since
the weights were 10 times lighter than the nails before immersion and they continue to
be 10 times lighter after immersion, the equilibrium will not be disturbed.

                                   * The figure wasn't given in the statement of the problem as the
                                exact share of the weight lost is immaterial here.
202-203                        Answers

                               A Piece of Soap
Three quarters of a piece of soap plus 3/4 kilogrammes weigh as much as the whole
piece. But a whole piece is 3/4 plus 1/4, hence 1/4 of a piece weighs 3/4 kilogrammes
and the whole piece weighs four times as much as 3/4 kilogrammes, i.e.
3 kilogrammes.
                               Cats and   Kittens
A comparison of both weighings shows that replacing a cat by a kitten reduces the
weight by 2 kilogrammes. It follows that a cat is 2 kilogrammes heavier than a kitten.
With this in mind we in the first weighing replace all the four cats by kittens to obtain
4 + 3 = 7 kittens that will together weigh not 15 kilogrammes but 2 x 4 = 8
kilogrammes less. Consequently, the seven kittens weigh 15 — 8 = 7 kilogrammes.
Hence a kitten weighs 1 kilogramme, and a cat weighs 1 + 2 = 3 kilogrammes.

                               Shell and Beads
Compare the first and second weighings. You'll see that in the first weighing the shell
can be replaced by one cube and eight beads. We'll then have four cubes and eight
beads on the left pan balanced by 12 beads. If we now remove eight beads from each
pan, we won't upset the balance. There'll four cubes now remain on the left pan, and
four beads on the right. One cube thus weighs the same as one bead.
  We can now work out the weight of the shell: replacing (second weighing) the cube
on the right pan by a bead gives that the weight of the shell is equal to that of nine
  The result can be checked easily.
  In the first weighing, replace the cubes and shell on the left pan by an appropriate
number of beads. You'll thus obtain 3 + 9 = 12, as required.
In the first weighing we replace one pear by six peaches and an apple. We may do so
because the pear weighs as much as the six peaches and apple. We then obtain four
apples and six peaches on the left pan and 10 peaches on the right. Removing the six
peaches from each pan gives that the four apples weigh as much as four peaches.
Accordingly one peach weighs the same as one apple.
  Now it's easy to figure out that a pear weighs the same as seven peaches.

                               How Many     Glasses?
The problem has several different solutions. The following is just one of them.
   In the third weighing we replace each jug by a bottle and a glass (we know from the
first weighing that the balance should remain in equilibrium). We thus find that two
bottles and two glasses are balanced by three saucers. It thus appears that four glasses
and two saucers are balanced by three saucers.
   Removing two saucers from each pan shows that four glasses are balanced by one
   Hence one bottle is balanced (compare with the second weighing) by five glasses.
                               9 Answers

 ***                           With a Weight and a Hammer
The procedure to be followed is like this. First put the hammer on one pan and the
weight on the other. Then add just enough sugar for the pans to be in equilibrium. It's
clear that the sugar weighs 900 — 500 = 400 grammes. The operation is performed
three more times. The remaining sugar weighs 2,000 — (4 x 400) = 400 grammes.
   It is only remains now to halve each of the five 400-gramme packets obtained. It's
a straightforward exercise: the contents of a 400-gramme packet are divided between
two packets put on different pans until the balance balances.

                              Archimedes's   Problem
If the crown ordered had been made purely of gold, it would have weighed
10 kilogrammes in air losing when immersed 1/20 part of its weight, i.e. 1/2
kilogramme. But we know that in fact the crown lost in water 1 0 - 9 1/4 = 3/4
kilogramme, not 1/2 kilogramme. This was because it contained silver-a metal that in
water loses 1/10 part of its weight, not 1/20 part. The crown thus contained an amount
of silver sufficient for it to lose in water 3/4 kilogramme, rather than 1/2 kilogramme,
i. e. 1 /4 kilogramme more. Suppose in the purely golden crown one kilogramme of gold
were replaced by silver, the crown would when immersed lose another 1/10-1/20 = 1/20
kilogramme. Consequently, in order to decrease the crown's weight by 1/4 kilogramme
it was necessary to replace with silver as many kilogrammes of gold as there were
1/20ths in 1/4: 1/4-4-1/20 = 5. So, the crown contained 5 kilogrammes of silver and
5 kilogrammes of gold instead of the 2 kilogrammes of silver and the 8 kilogrammes
of gold the craftsman was given. Thus, 3 kilogrammes of gold had been stolen and
replaced by silver.
204-205           Problems on Clocks and Watches

             72   Three Clocks
                  In my home there are three clocks. On the 1st of
                  January they all showed true time. But only the first
                  clock kept perfect time, the second was a minute slow
                  a day, and the third was gaining a minute a day.
                  Should the clocks continue like this, how long would it
                  take for them all to show true time again?

                  Two Clocks
                  Yesterday I checked my wall clock and alarm clock
                  and set them correctly. The wall clock is 2 minutes
                  slow an hour, the alarm clock gains 1 minute an hour.
                     Today both clocks stopped simultaneously since they
                  had run down. The wall clock shows 7 o'clock and the
                  alarm clock 8 o'clock.
                     At what time yesterday did I set the clocks?

                  What Time Is It?
                     "Where are you hurrying to?"
Figure 212           "To catch the 6 o'clock train. How long have I got
                     "50 minutes ago there were four times more minutes
                  after three."
                     What does this strange answer mean? What time was

                  When Do the Hands     Meet?
                  At 12 o'clock one hand is above the other But you
                  may have noticed that it is not the only moment when
                  the hands meet: they do so several times a day.
                    Can you say when all those moments are?
Figure 213

                  When are the Hands
                  Pointing in Opposite Directions?
                  By contrast, at 6 o'clock both hands point in opposite
                  directions. But is it only at 6 o'clock that this is the
                  case or there are some other such moments during the
                  next 12 hours?
             Problems on Clocks and Watches

             On Either Side of Six      O'Clock
Figure 214
             I glanced at a clock and noticed that both hands were
             equally separated from 6. What time was it?

             The Minute Hand Ahead of the Hour Hand
             When is the minute hand as far ahead of the hour hand
             as the hour hand in turn is ahead of the figure 12 on
             the face? And maybe there are several such moments
             during the day or none at all?

             Vice Versa
             If you observe a clock attentively, you may have
             noticed the reverse arrangement of the hands as
             compared with that just described: viz. the hour hand
             is as far ahead of the minute one as the minute hand is
             ahead of the figure 12.
                When does this happen?

             Three and Seven
             A clock strikes three. And while it does so 3 seconds
             elapse. How long does it take the clock to strike seven?
                I warn you, just in case, that this isn't a joke, i. e. it's
             not a trick question.

             Lastly, make a small experiment. Put your watch on
             a table, move a few steps aside and listen to the ticking.
             If it's sufficiently quiet in the room, you'll hear that
             your watch sounds, as it were, in intervals: ticks for
             a while, then is silent several seconds, and then starts
             ticking again, and so on.
206-207                        Answers

                               Three    Clocks

720 days. During this time the second clock will lose 720 minutes, i.e. exactly 12
hours, and the third clock will have gained exactly the same time. Then all the three
clocks will show as they did on the 1st of January, i.e. true time.

                               Two     Clocks
The alarm clock is gaining 3 minutes an hour compared with the wall clock. Thus it
gains an hour, i.e. 60 minutes, every 20 hours. But during these 20 hours the alarm
clock gains 20 minutes compared with true time. This implies that both clocks were
set correctly 19 hours 20 minutes before, i.e. at 11.40.

                               What Time Is It?
Between 3 and 6 o'clock there are 180 minutes. The number of minutes to go to
6 o'clock is easily found by dividing 180 — 50 = 130 minutes into two parts, one of
which being four times larger than the other. Hence, we'll have to find 1/5 part of 130.
It was thus 26 minutes to 6 o'clock.
   In fact, 50 minutes before it was 26 + 50 = 76 minutes to go to 6 o'clock.
Accordingly, 180 — 76 = 104 minutes had passed since 3 o'clock, which is four times
longer than the time to go to 6 o'clock.

                               When Do the Hands      Meet?
We start our observation at 12 o'clock, when both hands meet. Since the hour hand
moves 12 times slower than the minute one (it takes 12 hours to make a complete
circle, and the minute one 1 hour), the hands cannot, of course, meet during the next
hour. But after the hour has passed and the hour hand come to the 1 o'clock mark
(having completed 1/12th of the full circle), the minute hand has made a complete turn
and is again at 12, i.e. 1/12 part of the circle behind the hour hand. The condition of
the race is now different since the hour hand moves slower than the minute one, but is
ahead of the minute hand which has to overtake it. If the race lasted an hour, the
minute hand would have gone round a complete circle, and the hour hand 1/12 part
of the circle, i.e. the minute hand would have travelled 11/12 part of the circle more.
But to overtake the hour hand, the minute hand must only cover 1/12 of the circle
which is the distance separating them. This requires a period of time that is the same
fraction of an hour as 1 /12th is a fraction of 11/12 i.e. one eleventh. Thus, the hands
will meet in 1/11 hour, i.e. in 60/11 = 5 5/11 minutes.
   The hands will thus meet 5 5/11 minutes after the first hour has elapsed, i.e. at
5 5/11 minutes past one.
   What about the next meeting?
   You should be able to see that it'll occur 1 hour 5 5/11 minutes later, i.e. at 10
10/11 minutes past 2 o'clock. The next meeting occurs another 1 hour 5 5/11 minutes
later, i.e. at 16 4/11 minutes past 3 o'clock, and so forth. You may have already
guessed that, all in all, there'll be 11 such meetings. The 11th comes 1 1/11 x 11 = 12
                                9 Answers

hours after the first one, i.e. at 12 o'clock. In other words, it coincides with the first
meeting, with future meetings occurring at the previous times.
  Let's list the times of all the meetings:
                            1st - 5 5/11 minutes past    1 o'clock
                            2nd-10 10/11 minutes past 2 o'clock
                            3 r d l 6 4/11 minutes past  3 o'clock
                            4th-21 9/11 minutes past     4 o'clock
                            5th-27 3/11 minutes past     5 o'clock
                            6th—32 8/11 minutes past     6 o'clock
                            7th 38 2/11 minutes past     7 o'clock
                            8th—43 7/11 minutes past     8 o'clock
                            9th^49 1/11 minutes past     9 o'clock
                           10th-54 6/11 minutes past    10 o'clock
                           11th-                        12 o'clock

                                When Are the Hands Pointing in Opposite        Directions?
The approach here is very much like that in the previous problem. We'll again begin
at 12 o'clock when both hands meet. We want to find the time required for the minute
hand to get ahead of the hour hand by exactly half a circle, it is then that the hands
are pointing in opposite directions. We already know (see the previous problem) that
during an hour the minute hand gets ahead of the hour hand by 11/12 part of the
circle. For it to get ahead by only 1/2 a circle takes less than an hour by so rtiany
times as 1/2 is less than 11/12, i.e. 6/11 part of an hour. Accordingly, after 12 o'clock
the first time the hands point in opposite directions is in 6/11 hours, or 32 8/11
minutes. Look at a watch at this time and you'll see that the hands are really pointing
in opposite directions.
   Is this the only moment when we have such an arrangement? Of course, not. The
hands are so arranged 32 8/11 minutes after each meeting. We already know that
during a 12 hour's time there are 11 such meetings. Hence the hands point opposite
ways 11 times every 12 hours. These moments are easily found:
12 o'clock + 32 8/11 minutes = 32 8/11 minutes past 12 o'clock
  1 o'clock 5 5/11 minutes + 32 8/11 minutes = 38 2/11 minutes past 1 o'clock
  2 o'clock 10 10/11 minutes + 32 8/11 minutes = 43 7/11 minutes past 2 o'clock
  3 o'clock 16 4/11 minutes + 32 8/11 minutes = 49 1/11 minutes past 3 o'clock, and so
   I leave it for you to find the remaining moments.

                               On Either Side of Six        0'Clock
The problem is solved like the previous one. Imagine that both hands are at 12 and
that the hour hand has shifted by a certain part of a circle to be denoted by x.
Meanwhile the minute hand has turned by 12x. If the time that has passed is less than
one hour, then to meet the conditions of our problem the minute hand must travel
a full circle less the angle covered by the hour hand since 12. In other words,
                                1 - 12x = x.
  Hence 1 = 13x and x = 1/13 part of a circle. The hour hand covers this fraction of
a circle in 12/13 part of an hour, i.e. when it is (12/13) x 60 minutes or 55 5/13
minutes past 12 o'clock. During the same period of time, the minute hand covers 12
times more, i.e. 12/13 part of a circle. You see that both hands are equally separated
from 12, and hence equally separated from 6, too.
  We've found one location of the hands, namely one that comes about in the first
hour. During the second hour this occurs once more and you can find it arguing along
the same lines as before, from the relation
                               1 — (12x — 1) = x or 2 — 12x - x.
Hence 2 = 13x and x = 2/13 of a circle. So the hands will be in the right position at
(1 11/13) x 60 minutes or at 50 10/13 minutes past 1 o'clock.
   The hands will meet our requirement next time when the hour hand has shifted 3/13
of a circle away from 12, i.e. at 2 10/13 o'clock, and so on. AH told, there are 11 such
positions, the hands changing sides after 6 o'clock.

                               The Minute Hand Ahead of the Hour Hand
If we start looking at a clock at 12 o'clock exactly, then during the first hour we won't
see the position desired. Why? Because the hour hand covers 1/12 part of what the
minute hand does, and hence lags behind the minute hand far more than is required
for the arrangement we seek. Whichever the angle through which the minute hand
turns about 12, the hour hand will only be at 1/12 part of that angle, not a half as is
desired. But suppose an hour has elapsed and the minute hand is at 12 and the hour
hand at 1, i. e. 1/12 part of a complete turn ahead of the minute hand. Let's see if such
an arrangement of the hands may come about during the second hour. Suppose that
the moment has come when the hour hand has turned by a fraction of a circle that
we'll denote by x. Meanwhile the minute hand has covered 12 times more, i.e. 12x. If
now we subtract from this a complete turn, the difference 12x — 1 must be twice as
large as x, i. e. 2x. Thus 12x - 1 = 2x, whence it follows that a complete turn equals
lOx (because 12x — lOx = 2x). But if lOx equals a complete turn, then lx = 1/10 part
of a turn. We've thus arrived at the solution: the hour hand must have moved by 1/10
part of a turn past 12 o'clock. This takes 12/10 hours or 1 hour 12 minutes. The
minute hand will then be two times farther away from 12, i.e. at 1/5 of a turn away,
which corresponds to 60/5 = 12 minutes, as required.
   We've found one solution to the problem. But there are other ones and during
a period of 12 hours the hands come to be arranged in the right way several times.
We'll try to find the other solutions.
   To find the next time we'd have to wait till 2 o'clock and now the minute hand is at
12 and the hour hand at 2. Reasoning along the same lines as before we arrive at
                                  12x — 2 = 2x,
whence two complete turns are equal to lOx, and hence x = 1/5 part of a complete
turn. This corresponds to the moment 12/5 = 2 hours 24 minutes.
   I leave it to you to work out further moments. You'll find that the hands arrange
themselves in the right way at the following 10 instants in time:
                               1.12    7.12
                               2.24   8.24
14 - <>75
                               9 Answers

                               4.48      10.48
                               6.00      12.00
   The answers "6.00" and "12.00" might appear wrong, but only at first sight. In fact,
at 6 o'clock the hour hand is at 6 and the minute one, at 12, i. e. exactly twice as far.
At 12 o'clock the hour hand is separated from 12 by "zero" and the minute one, if you
wish, by "double zero" (because double zero is just zero). So, this case, too, meets the
restrictions of our problem.
                               Vice Versa
After the treatment we have just given, this problem is an easy exercise. Using the
same arguments as above we can determine that for the first time the required
arrangement will occur at the time given by
                             12x — 1 = x/2.
Therefore 1 = 11 1/2 x, or x = 2/23 of a turn, i.e. (2/23) x 12 hours or 1 1/23 hours
after 12 o'clock. Hence at 2 14/23 minutes past 1 o'clock the hands will be arranged
correctly. The minute hand will then be midway between 12 o'clock mark and 1 1/23
hours mark, i.e. at 12/23 hours mark, which is exactly 1/23 part of a turn (the hour
hand will be at 2/23 part of a turn).
   The hands will be arranged in the required manner for the second time at a time
which can be found from the relation
                               12x — 2 = x/2.
It follows that 2 = 11 1/2 x and x = 4/23, the time we seek is 5 5/23 minutes past
2 o'clock.
   The third moment is 7 19/23 minutes past 3 o'clock, and so forth.

                               Three and Seven
The commonest answer is "7 seconds". But, as we'll now see, that is wrong.
   When the clock strikes three we have two gaps: (1) between the first and second
strokes, (2) between the second and third strokes.
   Each gap thus lasts 1 1 / 2 seconds.
   But when the clock strikes seven, there are six such gaps, which gives 9 seconds.

The enigmatic interruptions in the ticking are only due to fatigue in your ears. From
time to time your perception of sound becomes blunted for a second or two so that in
these intervals you won't hear the ticking. This aural fatigue passes off after a short
while and previous ability to perceive the sound returns with the result that you again
hear the ticking. Then another fatigue period comes on, and so forth.
          Problems on Transport

          A Plane's Flight
          An aircraft covers the distance from town A to town
          B in 1 hour 20 minutes. However, it takes it 80 minutes
          to get back.
            How could you explain it?

          Two   Locomotives
          You may have seen a train driven by two locomotives,
          one at the front and the other at the back. But have
          you ever given any thought as to what happens to the
          couplings between the carriages and to their buffers?
          The front locomotive only pulls the carriages when the
          coupling is taut, in which case the buffers do not press
          against each other and so the rear locomotive cannot
          be pushing. On the other hand, when the rear
          locomotive pushes the train, the buffers press hard
          against each other, which makes the coupling become
          slack, thus rendering the front locomotive useless.
             It turns out that the locomotives cannot be moving
          the train at the same time since only one of them is
          working at a time.
             Why then do they employ two locomotives?

          The Speed of a Train
          You are travelling in a train and want to find its speed,
          could you work it out from the clatter of the wheels?

          Two   Trains
          Two trains once left their respective stations for the
          other's station simultaneously. The first arrived at its
          destination an hour after the two trains had met each
          other. The second reached its destination 2 hours 15
          minutes after the same event.
            How many times faster was the first train?
            The problem can be done using mental arithmetic.

          How Does a Train Start From Rest?
          You may have noticed that before making a train move
          forward the engine-driver sometimes makes it push
          back. Explain why.
Problems on Transport

A Race
Two sailing boats are competing against each other.
They must sail 24 kilometres there and back in the
shortest time possible. The first boat covered the whole
route with a uniform speed of 20 kilometres an hour
whilst the second boat sailed the outward leg at 16
kilometres an hour and sailed back at 24 kilometres an
   The first boat won, though it would seem that the
second one should have gained during the return trip
exactly what it lost out during the first section of the
route. It should thus have come in at the same time as
the first boat.
     Why did it lag behind?

Steaming Up and Down the River
A steamer makes 20 kilometres an hour downstream
and 15 kilometres an hour upstream. A trip between
two towns takes it 5 hours less than the return trip.
    What is the distance between the towns?
212-213                |       Answers

                                 A Plane's Flight
There is actually nothing to explain here, the aircraft takes the same time to travel
in both directions.
     The problem has a catch for the inattentive reader who might think that 1 hour
20 minutes and 80 minutes are different times. Strange as it might seem, many people
swallow this bait, and people used to adding up are more likely to make it than those
who aren't The explanation lies in the habit of dealing with metric system of measures
and money. We are apt to treat "1 hour 20 minutes" and "80 minutes" just like "1
pound 20 pence" and "80 pence", say, or "1 dollar 20 cents" and "80 cents". So it's
really a psychological problem.

                                 Two Locomotives
The way it works out is as follows. The front locomotive does not take care of the
whole of the train, but only part of it, about half the carriages. The rest of them are
pushed by the rear locomotive. The couplings between the first group of carriages are
taut whilst they are slack between the rear ones which are being pushed buffer to

                                   The Speed of a Train
You must have noticed that when travelling in a train you feel regular jerks all the
time which the springs, however good, cannot suppress. These jerks come from the
wheels being slightly jarred at rail junctions (Fig. 215) and are transferred throughout
the carriage.

Figure 215

     This nuisance, which is also bad both for the carriages and the tracks, lends itself
for measuring the speed of the train. You only need to count the number of jerks you
feel in one minute to find how many rails you've passed. Now you only have to
multiply this number by the length of a rail to arrive at the distance covered by the
train during that minute.
     The regulation length of a rail is about 15 metres*. So, multiply the number of
jerks a minute by 15 and then by 60, divide the result by 1,000 and you'll obtain the

                                  * You may work out the length of a rail by pacing it out, seven
                               paces amounting to about 5 metres.
                                9 Answers

number of kilometres covered by the train per hour. So,

                                  (number of jerks) x 15 x 60
                                                              = kilometres per hour.

                                  Two       Trains
The faster train arrives at the meeting point having covered a distance that is larger
than the distance covered by the slower train by as many times as the speed of the
faster train is higher than that of the slower train. After the meeting each train has to
pass the distance that had been covered by the other one. In other words, the faster
train covered a distance after the meeting that was as many times shorter than the
distance covered by the slower train as its speed was higher. If we denote the ratio of
the two speeds by x, then the faster train took x 2 times less time than the other to
cover the distance from the meeting point to the respective station. Hence x 2 = 2 1/4
and x = 1 1/2, i.e. the first train is 1.5 faster than the second train.

                                  How Does a Train Start From       Rest?
When a train arrives at a station and comes to rest, the couplings between the
carriages are taut. If the locomotive is to begin to pull the train like this, it would
have to start the whole of the train from rest at once, which might be too difficult
a task for it. On the other hand, if the locomotive first pushes the train backwards the
couplings are no longer taut and the train is started from rest carriage by carriage in
succession and that is much easier.
     In other words, the engine-driver does what a coachman does sometimes when
the coach is heavily loaded, i.e. he starts the coach and only then jumps on it,
otherwise the horse would have to push more load from rest.

                                  A Race
The second boat lagged behind because it travelled at 24 kilometres an hour for
a shorter time than it travelled at 16 kilometres an hour. In fact, it travelled at
24 km/h for 24/24 hours, i.e. 1 hour, and at 16 km/h for 24/16 hours, i.e. 1 1/2 hours.
Therefore, it lost more time on the journey "there" than gained on the way "back".

                                  Steaming Up and Down the River
Travelling downstream the steamer covers 1 kilometre in 3 minutes whilst travelling
upstream it covers 1 kilometre in 4 minutes. In the first case, the steamer gains
1 minute every kilometre, and as the total gain is 5 hours, or 300 minutes, the distance
between the towns is 300 kilometres.
                                  300        300     ^   _
Surprising Calculations

A Glass of Peas
Of course, you've seen peas many times and held
a glass in your hand, so that you must know the sizes
of these things. Imagine a glass filled to the brim with
dry peas. Thread all the peas on a piece of string like
     If the string is stretched, how long would it be

Water and Wine
One bottle contains a litre of wine, another a litre of
water. A spoonful of wine is transferred from the first
bottle into the second, and then a spoonful of the
mixture thus obtained is transferred from the second
bottle into the first one.
     What do we now have, more water in the first
bottle or more wine in the second?

A Die
Figure 216 shows a die, i.e. a cube with from 1 to
6 points on its six faces.
  Peter bets that if the cube is thrown four times in
succession, then it is bound to show 1 at least once.
   But Vladimir argues that the 1 will either not appear
at all with the four throws or it will show more than

  Who stands the better chance of winning?

The Yale Lock
The Yale lock was invented by an American, Linus
Yale, Jr., in 1865, and has come to be almost
universally used ever since. Despite its long history,
some people question the possibility of having a large
number of versions of the lock. But we need only to
look at the construction of the lock to see that it
provides for an almost unlimited number of variations.
   Figure 217 depicts the front view of the Yale lock.
You see a small circle around the key hole which is the
end face of the cylinder passing through the depth of
the lock. The lock opens when the cylmder turns, but
this is the crunch. The cylinder is secured by five short
             Surprising   Calculations

             steel pins (Fig. 217, right). Each pin actually consists of
             two pins, and the cylinder can only be turned when the
             double pins are so arranged that the cut lie at the
             boundary of the cylinder.
Figure 217

               The pins are arranged this way using a key with
             serrated edge. You just insert the key into the keyhole
             and the pins are lifted to the height required for the
             lock to open.
                You can easily see now that the number of the
             various combinations of heights in the lock can be
             exceedingly large. It depends on the number of ways in
             which each pin may be severed.
                Suppose that each pin may be divided into two parts
             in 10 ways only. Try and work out the number of
             combinations possible for the Yale lock.

             How Many          Portraits?
             Draw a portrait on a sheet of cardboard and cut it into
             several-say nine-stripes. Next draw other stripes
             showing various parts of the face so that any two
             neighbouring stripes belonging to different portraits
             might be fitted into another portrait without
             interrupting the lines. If you prepare, say, four stripes *
             for each part of the face, you'll have 36 stripes all in all.

                * These could be conveniently glued onto four faces of a square
Figure 218
216-217   Surprising   Calculations

          You'll now be able to make up a variety of faces by
          taking nine stripes each time.
             Shops once used to sell ready-made sets of these
          stripes (or blocks) to make up portraits (Fig. 218). It
          was claimed that of 36 stripes one could produce
          a thousand various faces.
             Is it so?
          Perhaps you can use the abacus and can set, say, 25
          pounds, on it. But the problem becomes more difficult
          if you must shift not seven beads, as usual, but 25
          beads. Just try.
             To be sure, nobody is going to do so in practice, but
          the problem is not intractable and the answer is rather

          Leaves of a Tree
          If we were to take all the leaves from an old tree, say
          a lime-tree, and place them side by side without any
          breaks, how long approximately would the line be?
          Would it be possible, for example, to encircle a large
          house with it?
          A Million      Steps
          You must know what a million is and can estimate the
          length of your step, so that you should easily be able to
          say how far would a million steps take you? More than
          10 kilometres away? O r less?

          Cubic    Metre
          A teacher asked his class if they were to put all the
          millimetre cubes contained in cubic metre, on top of
          each other, how high would the column be?
            "It'd be higher than the Eiffel Tower (300 metres)!"
          one student exclaimed.
            "Even higher than Mont Blanc (5 kilometres)!"
          another answered. Which was closer to the truth?

          Whose Count Was             Higher?
          Two people kept count of the passers-by on
          a pavement over a period of an hour. One of them
          stood near the gate of a house whilst the other strolled
          to and fro along the pavement.
             Whose count was higher?

                      w        A Glass of Peas
Any guess here will lead you to an error. A calculation, however crude, is in order.
  A dry pea is about 1/2 centimetre across. A centimetre cube contains no less than
2 x 2 x 2 = 8 peas (if tightly packed, even more). In a glass of capacity 250 cubic
centimetres there are no less than 8 x 250 = 2,000 peas. When strung these would give
a line 1/2 x 2,000= 1,000 centimetres long, i.e. 10 metres.

                               Water and Wine
In solving the problem we mustn't overlook the fact that the final volume of liquid in
the bottles was equal to the initial one, 1 litre. We then argue as follows. Let, after
both transferrals, the second bottle contain n cubic centimetres of wine, and hence
(1,000 - n) cubic centimetres of water. Where have the missing n cubic centimetres of
water gone? Clearly, these are to be found in the first bottle. Accordingly, the wine in
the end contains as much water as there is wine in the water.

                               A Die
 The number of all the possible events after four throws of the die is 6 x 6 x 6 x 6 =
 = 1,296. Suppose that the die has already been thrown once and a 1 appeared. Then
for the three remaining throws the number of all the possible events, favourable for
Peter, i.e. the occurrence of any face save for the 1, will be 5 x 5 x 5 = 125. In exactly
the same way, 125 outcomes favourable for Peter are possible if the 1 appears only in
the second, only in the third or only in the fourth throw. So, there are 125 + 25 +
 + 125 + 125 = 500 various possibilities for the 1 to appear once, and only once, in the
four throws. As to the unfavourable outcomes, there are 1,296 — 500 = 796 since all
the remaining events are unfavourable.
  Thus, we see that Vladimir stands better chance to win than Peter: 797 against 500.

                               The Yale Lock
It's easily seen that the number of different locks possible is 10 x 10 x 10 x 10 x 10 =
 = 100,000. Each of these locks can only be opened by their own key. It is very
comforting for the owner of the lock that there are 100,000 versions of the lock and
key as the lock picker has only one chance in 100,000 to hit upon the right key.
   Our calculation is very rough since it assumes that each pin can only be divided in
10 different ways. Clearly it could actually be done in a larger number of ways, thus
notably increasing the number of different locks possible. This shows the advantage of
the Yale lock.

                               How Many      Portraits?
Far more than a thousand. To show this is true make each of the nine sections of
a portrait with one of the Roman numerals I, II, III, IV, V, VI, VII, VIII and IX. For
each section there are four stripes, so we'll mark these by 1, 2, 3 and 4.
    Take stripe 1.1. It may go to II. 1, II.2, II.3, or II.4, i.e. we may have four
combinations. But since section I may be represented by four stripes (1.1, 1.2, 1.3, or
1.4) each of which may be connected to II in four ways, then the two upper sections
I and II may be joined in 4 x 4 = 16 various ways.
    To each of these 16 arrangements we may attach section III in four ways (III.l,
 III.2, III.3, or III.4). Consequently, the first three sections may be combined in 16 x
  x 4 = 64 various ways.
    Reasoning along the same lines, we find that I, II, III, and IV may be arranged in
64 x 4 = 256 various ways: I, II, III, IV, and V, in 1,024 ways: I, II, III, IV, V, and VI,
 in 4,096 ways, and so forth. Lastly, all the nine sections may be fitted together in 4 x
  x 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4 = 262,144 ways. Not one thousand but more than
a quarter of a million different portraits!
    The problem is a very instructive one and it goes to explain why it's only
exceptionally rarely that we may come across two similar faces. We've just seen that if
the human face were characterized by as few as nine features with only four versions
possible, then it would be more than 260,000 various faces in existence. In actuality,
there are more than nine features and they may vary in more than four ways. In fact,
if there were 20 features varying in 10 ways each, we would have 1020, or
100,000,000,000,000,000,000, ways.
    Incidentally, this is many times greater than the world's population.

You can set 25 pounds using 25 beads in the following way:

Fiaure 219

Actually, this gives 20 pounds + 4 pounds + 90 pence + 10 pence = 25 pounds.
  The number of beads is 2 + 4 + 9 + 10 = 25.

                                Leaves of a Tree
A small town, let alone a house, could be encircled with the leaves from a tree if we
arranged them in a line because the line would be about 12 kilometres long! Really,
the foilage of mature tree includes no less than 200-300 thousand leaves. If for
definiteness we stick to 250 thousand and take a leaf to be 5 centimetres wide, we'll
have a line 1,250,000 centimetres long, which is 12,500 metres, or 12.5 kilometres.
 9                            Answers

                              A Million   Paces
A million paces is much more than 10 or even 100 kilometres. If an average pace is
about 3/4 metre long, then 1,000,000 paces = 750 kilometres. Since the distance from
Moscow to Leningrad is about 640 kilometres, then a million paces would take farther
than Leningrad.

                              A Cubic Metre
Both answers are far from the true figure because the column would be 100 times
higher than the highest mountain on Earth. Indeed, in a cubic metre there are 1,000 x
 x 1,000 x 1,000 = 1 milliard cubic millimetres. If you put one on top of another they
would form a column 1,000,000,000 millimetres high or 1,000,000 metres, or 1,000

                              Whose Count Was      Higher?
The counts were equal.
          Instructor and Student
          The story related below is said to have occurred in
          Ancient Greece. The teacher and thinker Protagoras
          (485-410 B.C.) undertook to teach a young man the art
          of being a barrister. The two sides made a deal that the
          student pay the fee just after he had made some
          achievement, i. e. after he had won his first trial.
             The student passed the course and Protagoras was
          waiting for his reward, but the student wouldn't appear
          in a court of justice. What was to be done? To get his
          fee the teacher sued his student. He argued thus: if he
          won the case, the money would be recovered by the
          court, whereas if he lost the case, and hence his student
          won it, the money would again be paid according to
          their deal.
             The student, however, regarded Protagoras's case as
          absolutely hopeless (he seems to have learned
          something from his teacher) and reasoned as follows: if
          the judge decided against him, he wouldn't pay
          according to the terms of the deal since he would have
          lost his first case, whereas if the judge decided in his
          favour, again he wouldn't have to pay since that would
          be the decision of the court.
             The judge was embarrased but after a great deal of
          thought he hit upon an idea and passed a decision that,
          without violating the terms of the deal, gave the teacher
          an opportunity to recover his fee.
             What was the decision?
          The   Legacy
          Here is another ancient problem that was a favourite
          with lawyers in Ancient Rome.
            A widow has to share a legacy of 3,500 sestertii with
          her child who was about to be born. According to
          Roman law, if the child were a boy, his mother got
          a half of the son's share but if it were a girl, the mother
          got double the share of the daughter. But it so
          happened that twins were born, a boy an a girl.
            How was the legacy to be shared so that the law was
          completely satisfied?
          Consider of jug containing 4 litres of milk. The milk
          must be divided equally between two friends, but the

only containers available are two empty jugs, one of
which holds 2 1/2 litres and the other holds 1 1/2 litres.
  How can the milk be divided using the three jugs?
  Of course, it'll be necessary to pour the milk from
one jug into another. But how?

Two     Candles
The electricity failed in my flat because the fuse had
blown. I lighted two candles that had been specially
prepared on my desk, and worked on in their light
until the failure was set right.
   The next day they wanted to know how long the
electricity was off. I had not noticed the time when the
electricity failed and was restored, and I didn't know
the initial length of the candles. I only knew that the
candles were the same length but different thicknesses,
and that the thicker one took 5 hours to burn down
completely whilst the thinner one took 4 hours. Both
were new before I had lighted them up. But I didn't
find the ends of the candles: somebody had thrown
them away. I was told that the stubs were so small that
it wouldn't have paid to keep them.
   "But couldn't you remember their lengths?" I asked.
   "They weren't the same. One was four times longer
than the other".
   All my attempts to squeeze out something more
failed. I had to be content with the above information
and try to work out how long the candles had been

   How would you handle the problem?

Three Soldiers
Three soldiers were having a problem, too. They had to
cross a river without a bridge. Two boys with a boat
agreed to help the soldiers but the boat was so small it
could only support one soldier and even then a soldier
and a boy couldn't be in the boat for fear of sinking it.
None of the soldiers could swim.
  It would seem that under these conditions only one
soldier could cross the river. However, all three soldiers
were soon on the other bank and returned the boat to
the boys.
  How did they do it?
222-223   Predicaments

          A Herd of Cows
          Here is one of the versions of a curious ancient
            A father distributed his herd amongst his sons. To
          his eldest he gave one cow plus 1/7 of the remaining
          cows; to his second eldest, two cows plus 1/7 of the
          remaining cows; to the third eldest, three cows plus 1/7
          of the remaining cows; to the fourth eldest, four cows
          and 1/7 of the remaining cows, and so forth. The herd
          was distributed among his sons without remainder.
             How many sons and how many cows were there?

          Square     Metre
          When a boy was told for the first time that a square
          metre contains a million square millimetres, he
          wouldn't believe it.
            "Why so many?" he was surprised. "Here I've got
          a sheet of graph paper that is exactly one metre long
          and one metre wide. And are there million millimetre
          squares here? I don't believe it!"
            "Count them then," somebody advised.
            The boy decided to do so and count all the squares.
          He got up early in the morning and set about counting
          them neatly marking each square he had counted with
          a point.
            Each mark took him a second so the going was rather
          fast. He worked like blazes, still do you think he
          managed to make sure that a square metre has
          a million square millimetres on the same day?

          A Hundred        Nuts
          A hundred nuts are to be divided between 25 people so
          that nobody gets an even number of nuts.
            Could you do it?

          Dividing       Money
          Two people were making porridge on a camp-fire. One
          contributed 200 grammes of cereals, the other 300
          grammes. When the porridge was ready, they were
          joined by a passer-by who partook of their meal and
          paid them 50 pence.
             How should they divide the money?

Sharing Apples
Nine apples must be shared out amongst 12 children so
that no apple is divided into more than four parts. On
the face of it the problem is insolvable, but those who
knows about fractions can solve it easily.
   Once you have solved that one it should be easy to
handle another problem in the same vein: to divide
seven apples among 12 boys so that none of the apples
is divided into more than four parts.

A Further Apple     Problem
Five friends came to see Peter. Peter's father wanted to
treat all six boys to apples but there were only five
apples. What was to be done? Everyone had to have
his fair share. The apples, of course, had to be cut but
not into small pieces since Peter's father wouldn't cut
them into more than three. So, the problem was to
divide the five apples equally among the six boys so
that none of the apples was cut into more than three
pieces. How was Peter's father to get out of his

One Boat for     Three
Three sports enthusiasts possess one boat. They keep it
on a chain with three locks so that each of them could
use it but a stranger couldn't. Each of them has his
own key but he can still unlock the boat without
waiting for his friends and their keys.
  How did they arrange it?

Waiting for a Tram
Three brothers came to a tram stop. There was no tram
in sight and the eldest brother suggested they wait.
   "Why wait?" the second brother asked, "we'd better
go on. When the tram catches up with us, we can jump
onto it, but by then we'd have got part of the way
home and thus we'll get there sooner."
   "If we decide to go," the youngest brother objected,
"then we'd better go backwards not forwards: since
then we'll meet an oncoming tram sooner and so get
home sooner."
  Since the brothers couldn't persuade each other, each
went his own way. The eldest stayed to wait, the second
went on, and the youngest walked back down the
route. Which of the three got home sooner? Who was
the most reasonable?
224-225                        Answers

                               Instructor and Student

The decision was to decide against Protagoras but give him the right to bring the case
before the court a second time. After the student had won his first trial, the second
one should undoubtedly be decided in favour of the instructor.

                               The       Legacy
The widow gets 1,000 sestertii, the son 2,000 sestertii, and the daughter 500 sestertii.
This fulfils Roman law since the widow gets a half of the son's share and double the

Seven pourings will be required as is shown in the table:

                               Pouring            41         11/21       21/21

                               1                  11/2       —           2 1/2
                               2                  11/2       1 1/2       1
                               3                  3          —           1
                               4                  3          1           —
                               5                       1/2   1           2 1/2
                               6                       1/2   1 1/2       2
                               7                  2          —            2

                               Two Candles
We'll construct a simple equation. We'll denote the time (in hours) that the candles
burned by x. Each hour 1/5 part of the original length of the thick candle and 1/4 part
of the original length of the thin candle burns away. Accordingly, the thick candle's
stub will be 1 - x/5 of its original length and the thin candle's stub 1 - x/4 of the
original length. We know that the candles were originally equally long and that the
four times the length of the thick stub, i.e. 4(1 - x/5), was equal to the length of the
thin stub (1 - x/4). Thus,

Solving the equation gives that x = 3 3/4 hours, i. e. the candles had burned for
3 hours 45 minutes.

                               Three Soldiers
The following six crossings were made:
1st crossing. Both boys go to the opposite bank and one of them brings the boat back
to the soldiers (the other stays on the opposite bank).
9                              Answers

2nd crossing. The boy that brought the boat back stays on the bank with the soldiers
and a soldier crosses the river in the boat. The boat returns with the other boy.
3d crossing. Both boys cross the river and one of them returns with the boat.
4th crossing. The second soldier crosses and the boat returns with the boy.
5th crossing. Like the third one.
6th crossing. The third soldier crosses and the boat returns with the boy. The boys
continue on their journey and the three soldiers are on the opposite bank.

                              A Herd of Cows
Arithmetically (i.e. without resorting to equations), the problem should be approached
from the end.
   The youngest son got as many cows as there were sons for he could not get an
additional 1/7 of the remaining herd as there were no cows left.
   Further, the next son got one cow less than there were sons, plus 1/7 of the
remaining cows. Accordingly, the share of the youngest son amounts to 6/7 of the
share of the remainder.
   It thus follows that the number of cows the youngest son got must be divisible by
   Let's assume that the youngest son received six cows and see if this assumption is
good. It follows from the assumption that there were six sons. The fifth son got five
cows plus 1/7 of seven, i.e. six cows all in all. Thus, the two youngest sons got 6 +
 + 6 = 1 2 cows, which accounts for 6/7 part of the herd left after the fourth son has
received his share. The total residue was 12^-6/7 = 14 cows, hence the fourth son got
4 + 14/7 = 6 cows.
   We'll now work out the residue after the third son got his share: 6 + 6 + 6 = 18
cows is 6/7 part of the residue. Therefore, the total residue was 18—:—6/7 = 21 cows.
The third son got 3 + 21/7 = 6 cows.
   In exactly the same way we'll find that the second and first sons also got six cows
   Our assumption that there were six sons and 36 cows appears to be plausible.
   But are there other solutions? Assume that there were 12 sons, not six. It turns out
that this assumption is unsuitable. The number 18 won't do either. Other multiples of
six would be unreasonable since there couldn't be 24 or more sons.

                              Square     Metre
No, the boy would not be able to verify the fact in one day. Even if he counted for 24
hours without interruption, he would have counted only 86,400 squares since there are
only 86,400 seconds in 24 hours. To count to one million he would have to work for
almost 12 days without stopping, and for a month if he worked 8 hours a day.

                              A Hundred      Nuts
Many people would immediately set about tiying a variety of combinations, but their
efforts would all be to no avail. If you give some thought to the problem, you'll
understand the futility of all their efforts since the problem is insolvable.
   If you could break 100 into 25 odd summands, you would have been able to make
an odd number of odd numbers add up to 100 which is an even number, and that is
clearly impossible.
   In fact, we would have to obtain 12 pairs of odd numbers and one more odd
number. Each pair of odd numbers yields an even number, so 12 pairs of even
numbers must add up to an even number. If then we add an odd number to the total,
we'll end up with an odd result. Thus 100 can never be composed of such summands.

                               Dividing   Money
Most people answer that the one who contributed the 200 grammes should get 20
pence and the other 30 pence. This division is not fair.
   We'll argue as follows: 50 pence was paid for one portion of food.
   Since there were three eaters, the cost of the porridge (500 grammes) should be
1 pound 50 pence. The person who contributed the 200 grammes gave 60 pence worth
o f food in terms of money (since a hundred grammes costs 150 :- 5 = 30 pence).
However, he also consumed 50 pence worth of porridge, hence he must get back 60 —
 — 50 = 10 pence.
   The contributor of the 300 grammes (i.e. 90 pence in terms of money) must get
90 - 50 = 40 pence.
   Thus, out of the 50 pence one person should have 10 pence and the other person 40

                               Sharing Apples
It's possible to share nine apples equally between 12 children without cutting any
apple into more than four parts.
   Six apples should be divided in two each to yield 12 halves. The remaining three
apples should each be divided into four equal parts to yield 12 quarters. Now each
child receives a half and a quarter. So each will get 3/4 of an apple as required,
because 9-4-12 = 3/4.
   Reasoning along the same lines- it's possible to divide seven apples among 12
children so that each child gets an equal share and no apple needs to be cut into more
than four parts. In this case each child should get 7/12 of an apple, but notice that
7/12 = 3/12 + 4/12 = 1 / 4 + 1/3.
   Therefore three apples are divided into four parts and the four remaining apples
into three parts each. We thus obtain 12 quarters and 12 thirds.
   In consequence, each child can be given a quarter and a third, or 7/12.

                               A Further Apple     Problem
The apples were divided thus: three apples were each cut in half to yield six halves
that were distributed among the children and the remaining two apples were each cut
into three to obtain six thirds that were also given to the children.
   Consequently, each boy got a half and a third of an apple, i. e. all the boys got their
equal share, and none of the apples was cut into more than three equal parts.
                              9 Answers

'5 Q0                         One Boat for   Three
The locks should be connected as shown in Fig. 220. You can see quite easily that
each of the boat's owners can open the chain of the three locks using his key.

Figure   220

                              Waiting for a Tram
The youngest brother, who went backwards, saw an oncoming tram and jumped into
it. When the tram came to the stop where the eldest brother was waiting, he got in
too. A short while later the tram caught up the third brother who was walking
homewards and collected him. All the three brothers found themselves in the same
tram and, of course, arrived home at the same time.
   The most reasonable brother was the eldest one since he waited quietly at the stop.
228-229   Problems from Gulliver's          Travels

          Beyond doubt the most fascinating pages in Gulliver's
          Travels are those describing his unusual adventures in
          the country of tiny Lilliputians and in the country of
          giant Brobdingnagians. In Lilliput the dimensions-
          height, width, thickness-of people, animals, plants and
          other things were 1/12 of those here. By contrast, in
          Brobdingnag they were 12 times larger. We can easily
          understand why the author of the Travels choose the
          number 12, if we remember that in the British system of
          units there are 12 inches in a foot. A 12-fold increase or
          decrease doesn't seem to be very much of a change but
          the nature and way of life in this fantastic countries was
          strikingly different from those we are used to. Every
          now and then the differences are so amazing that can
          serve as a material for interesting problems.

          Animals of Lilliput
          Gulliver relates: "Fifteen hundred of the Emperor's
          largest horses... were employed to draw me towards
          the metropolis."
             Doesn't it seem to you that 1,500 horses are a bit too
          many taking into account the relative dimensions of
          Gulliver and Lilliputian horses?
             Also, Gulliver tells us a no less amazing thing about
          the cows, bulls, and sheep, for when he left he just "put
          them into his pocket".
             Is it all possible?

          Hard Bed
          Lilliputians made the following bed for their giant
          guest: "Six hundred beds of the common measure were
          brought in carriages, and worked u p in my house;
          a hundred and fifty of their beds sewn together made
          up the breadth and length, and these were four double,
          which however kept me but very indifferently from the
          hardness of the floor, that was of smooth stone".
            Why was Gulliver so incomfortable on the bed?
            And is this computation correct?

          Gulliver's Boat
          Gulliver left Lilliput in a boat washed up on the shore
          by chance. The boat seemed monstrous to the
             Problems from Gulliver's   Travels

             Lilliputians, it surpassed by far the largest ships of their
                Could you work out the displacement* of the boat
             in Lilliputian tonnes if its weight-carrying capacity was
             300 kilogrammes?

             Hogsheads and Buckets of Lilliputians
             Gulliver is drinking:
             "I made another sign that I wanted drink . They slung
             up with great dexterity one of their largest hogsheads;
             then rolled it towards my hand, and beat out the top;
Figure 221   I drank it off at a draught, which I might well do, for it
             hardly held half a pint... They brought me a second
             hogshead, which I drank in the same manner, and
             made signs for more, but they had none to give to me".
                Elsewhere in the book Gulliver describes the
             Lilliputian buckets as being no larger than a thimble.
                Why should such tiny hogsheads and buckets exist in
             a country where everything is only 1/12th normal size?
             Food Allowance and Dinner
             Lilliputians set the following daily allowance of food
             for Gulliver:
             "... the said Man Mountain shall have a daily
             allowance of meat and drink, sufficient for the support
             of 1,728 of our subjects."
                Elsewhere Gulliver relates:
             "I had three hundred cooks to dress my victuals, in
             little convenient huts built about my house, where they
Figure 222   and their families lived, and prepared me two dishes
             apiece. I took up twenty waiters in my hand, and
             placed them on the table; a hundred more attended
             below on the ground, some with dishes of meat, and
             some with barrels of wine and other liquors slung on
             their shoulders; all which the waiters above drew up as
             I wanted, in a very ingenious manner, by certain cords,
             as we draw the bucket up a well in Europe".
                 How did they come to fix on that number? And
             what is the use of all that army of servants to feed just
             one man? After all, he's only a dozen times taller than
             a Lilliputian. Are the allowance and appetites
             compatible with the relative sizes of Gulliver and the
             Lilliputians ?
                * The displacement of a ship is the largest load (including the
             weight of the ship itself) that the ship can support.
230-231      Problems from Gulliver's   Travels

             Three Hundred Tailors
Figure 223
             "Three hundred tailors were employed... to make me
                Was this army of tailors really necessary to have
             clothes made for a man who is only a dozen times
             larger than a Lilliputian?

             Gigantic Apples and Nuts
             In the part "A Voyage to Brobdingnag" devoted to
             Gulliver's stay in the country of giants we read about
             some of the hero's trouble-filled adventures. So once he
             was in the gardens of the court under some apple-trees
             and the Queen dwarf "when I was walking under one
             of them, shook it directly over my head, by which
             a dozen apples, each of them near as large as a Bristol
             barrel, came tumbling about my ears; one of them hit
             me on the back as I chanced to stoop, and knocked me
             down flat on my face."
                On another occasion "an unlucky schoolboy aimed
             a hazelnut directly at my head, which very narrowly
             missed me; otherwise, it came with so much violence,
             that it would have infallibly knocked out my brains; for
             it was almost as large as a small pumpkin".
                What do you think was the weight of the apples and
             nuts in Brobdingnag?
             A Ring of the Giants
             The collection of rarities brought by Gulliver from
             Brobdingnag includes "a gold ring which one day she
             (the Queen) made me a present of in a most obliging
             manner, taking it from her little finger, and throwing it
             over my head like a collar."
                Is it possible that a ring from a little finger would fit
             on Gulliver like a collar and how much, approximately,
             would the ring weigh?

             Books of the Giants
             About books of Brobdingnagians Gulliver tells us the
             following: "I had liberty to borrow what books
             I pleased. The Queen's joiner had contrived... a kind of
             wooden machine five and twenty foot high, formed like
             a standing ladder; the steps were each fifty foot long. It
             was indeed a movable pair of stairs, the lowest end
             placed at ten foot distance from the wall of the
Problems   from   Culliver's   Travek

chamber. The book I had a mind to read was put up
leaning against the wall. I first mounted to the upper
step of the ladder, and turning my face towards the
book, began at the top of the page, and so walking to
the right and left about eight or ten paces according to
the length of the lines, till I had gotten a little below
the level of my eye; and then descending gradually till
I came to the bottom; after which I mounted again,
and began the other page on the same manner, and so
turned over the leaf, which I could easily do with both
my hands, for it was as thick and stiff as a pasteboard,
and in the largest folios not above eighteen or twenty
foot long."
   Does this make sense?

Collars for the Giants
Finally, consider a problem of this kind that is not
directly taken from Gulliver's Travels.
   You may know that the size of a collar is nothing
but the number of centimetres of its length. If your
neck is 38 centimetres round, your collar size is 38. On
average an adult's neck is 40 centimetres round.
   If Gulliver wished to order some collars in London
for a Brobdingnagian, what number would he require?
232-233                         Answers

                    i6         Animals of Lilliput
It's calculated in the answer to "Food Allowance and Dinner" that Gulliver's volume
was 1,728 times larger than that of a Lilliputian. Crearly, he was that many times
heavier. For Lilliputians it was as difficult to transport his body as it would have been
to transport 1,728 grown-up Lilliputians. That is why the cart with Gulliver had to be
pulled by so many Lilliputian horses.

Figure 224

   Animals in Lilliput were also 1,728 times smaller in volume, and hence as much
 ighter than ours.
   Our cow is about 1.5 metres high and weighs 400 kilogrammes. A cow in Lilliput
would be 12 centimetres high and weigh 400/1,728 kilogrammes, i.e. less than 1/4
kilogrammes. A toy cow like this really could be carried about in a pocket.
   Gulliver gives a true account of relative sizes:
   "The tallest horses and oxen are between four and five inches in height, the sheep
an inch and a half, more or less; their geese about the bigness of a sparrow, and so
the several gradations downwards, till you come to the smallest, which to my sight
were almost invisible... I have been much pleased with observing a cook pulling
a lark, which was not so large as a common fly; and a young girl threading an
invisible needle with invisible silk."

                               Hard Bed
The calculation is quite correct. If a Lilliputian bed is 12 times shorter, and of course
12 times narrower than a conventional bed, then its surface "area would be 12 x 12
times smaller than the surface of our bed. Accordingly, for his bed Gulliver required
144 (i.e. to make a round number, about 150) Lilliputian beds. The bed would
however have been exceedingly t h i n - 1 2 times thinner than ours. Thus even four
layers of such beds would not have been soft enough for Gulliver since the resultant
mattress was three times thinner than ours.
                                Gulliver's   Boat
We know from the question that the boat could carry 300 kilogrammes, i.e. its
displacement was about 1/3 tonne. A tonne is the weight of 1 cubic metre of water,
hence the boat displaced 1/3 of our cubic metre. But all the linear dimensions in
Lilliput are 1/12 of ours, and volumes are 1/1,728 of ours. So 1/3 of our cubic metre
contains about 575 Lilliputian cubic metres and thus Gulliver's boat had
a displacement of 575 tonnes or thereabout since we arbitrarily took the figure 300
   Today we have ships with displacements of tens of thousands of tonnes ploughing
the seas, so a ship with a 575-tonne displacement should not be a wonder. We should
remember though that at the time of writing (early in the 18th century) 500-600-tonne
ships were still rare.

                                Hogsheads and Buckets of         Lilliputians
Lilliputian vessels were 12 times smaller than ours in every dimensions-height, width,
and l e n g t h - a n d 1,728 times smaller in volume. If we assume that our bucket contains
about 60 glasses, we can work out that a Lilliputian bucket contains only 60/1,728, i.e.
about 1/30 of a glass. This is just larger than a tea-spoonful but not really much larger
than the volume of a large thimble.
  If the capacity of a Lilliputian bucket is thus a tea-spoonful, the capacity of
a 10-bucket hogshead would not be much larger than half a glass. N o wonder
Gulliver couldn't quench his thirst with two such hogsheads.

                                Food Allowance      and Dinner
The computation is perfectly correct. We shouldn't forget that Lilliputians were an
exact, though smaller, replica of conventional people with normally proportioned
members. Consequently, they were not only 12 times shorter, but also 12 times
narrower and 12 times thinner than Gulliver, and their volume was 1/1,728 of that of
Gulliver. And to support the life of such a body requires respectively more food.
That's why Lilliputians calculated that Gulliver needed an allowance sufficient to
support 1,728 Lilliputians.
  We now see the purpose of so many cooks. To make 1,728 dinners requires no less
than 300 cooks taking that one Lilliputian cook can make half a dozen Lilliputian
dinners. An accordingly larger number of people is required to haul the load up to
Gulliver's table, which can be estimated to be the height of a three-storey building in

                                Three Hundred       Tailors
The surface of Gulliver's body was 1 2 x 1 2 , i.e. 144 times larger than that of
a Lilliputian. This is clearer if we imagine that each square inch of the surface of
a Lilliputian's body corresponds to a square foot on the surface of Gulliver's body.
We know, however, that there are 144 square inches in a square foot. Thus Gulliver's
suit would take 144 times more fabric than that of a Lilliputian, and hence more
234-235                        Answers

working time. If, say, one tailor can make one suit in two days, then to make 144 suits
in a day (or one of Gulliver's suits) may require 300 tailors.

                               Gigantic Apples and Nuts
An apple that weighs about 100 grammes here should correspond to an apple in
Brobdingnag that is as many times heavier as it is bigger in volume, i.e. 1,728 times
heavier than here. Thus Brobdingnagian apples are about 173 kilogrammes. If such an
apple falls from a tree and hits a man on the back, he would only just survive the
blow. Gulliver thus got off lightly.
Figure 225

   A Brobdingnagian nut must have weighed 3-4 kilogrammes, if we take that our nut
weighs about 2 grammes. Such a gigantic nut might be about a dozen centimetres
across. A 3-kilogramme, hard object thrown with the speed of the nut clearly could
smash the skull of a normal-size man. Elsewhere in the book Gulliver recalls: "There
suddenly fell such a violent shower of hail, that I was immediately by the force of it
struck to the ground: and when I was down, the hailstones gave me such cruel bangs
all over the body, as if I had been pelted with tennis balls." Quite plausible, because
each piece of hail in this country of giants must weigh no less than a kilogramme.

                               A Ring of the Giants
A normal little finger is about 11/2 centimetres across. Multiplying this by 12 gives 18
centimetres and a ring of such a diameter has a circumference of 56 centimetres, i. e.

Figure 226

it's sufficiently large for a normal head to go through it.
    As to the weight of such a ring, if a normal ring weighs 5 grammes its counterpart
in Brobdingnag must have weighed 8 1/2 kilogrammes!

                               Books of the Giants
If we start from the size of books current in our times (about 25 centimetres long and
12 centimetres wide), then Gulliver's account might appear to be a slight exaggeration.
Figure 227

You could handle a book 3 metres high and 1 1/2 metres wide without a ladder and
without having to move to the left or right by 8—10 steps. In the days of Swift, early in
the 18th century, the conventional format of books (tomes) was far larger than now.
20 x 30 cm formats were not uncommon, which when multiplied by 12 gives 360 x
 x 240 centimetres. It is impossible to read a 4-metre book without a ladder. But
a real tome of the time might be as large as a newspaper.
   However, the modest tome we mentioned would in the country of giants weigh
1,728 times more than here, i.e. about 3 tonnes. Assuming that it has 500 sheets, each
of its sheets would weigh about 6 kilogrammes, perhaps a bit too much for fingers.

                               Collars for the Giants
The neck of a giant will be 12 times larger than that of a normal man. And if
a normal man needs a collar of size 40, the giant would need a 40 x 12 = 480 size
                                         * * *

  We thus see that all the whimsical things in Swift seem to have been carefully
calculated. Responding to certain critisisms of his poem Eugine Onegin Alexander
Pushkin once noted that in his book "time is calculated with a calendar". In exactly
the same way Swift could say that all his objects had conscientiously been computed
using the laws of geometry.
236-237   Stories about Giant Numbers

          According to legend, the following happened in ancient

          I. General Terentius had returned to Rome with booty
          after a victorious campaign. Back in the capital he was
          received by the Emperor.
             The reception was very warm and the Emperor
          thanked him cordially for his services to the Empire
          promising to confer on him a high office in the Senate.
             But Terentius didn't want this. He said:
             "I have won many victories to exalt your grandeur,
          Sire, and to cover your name with glory. I have been
          unafraid of death and if I had many lives, I'd sacrifice
          all of them to you. But I'm tired of fighting, my youth
          had passed and my blood flows slower in my veins. The
          time has come for me to retire to my father's home and
          revel in the joys of domestic life."
             "What would you like to receive from me,
          Terentius?" the Emperor asked.
             "Hear me out with indulgence, Sire! In all these long
          years of battle, imbruing my sword with blood, I have
          had no time to take care of my well-being. I'm poor,
             "Proceed, brave Terentius."
             The encouraged general went on to say: "If it is your
          desire to reward your humble servant, then may your
          generosity help me live out the remainder of my days in
          peace and comfort at home. I do not seek honour or
          high office as I would like to retire from power and
          public life to live peacefully. Sire, please award me with
          money to provide for the rest of my life."
             The E m p e r o r - s o the legend goes-wasn't distingui-
          shed for his lavishness. He liked to save money for him-
          self but was miserly with it to others. The general's
          request plunged him in a deep reverie.
             He asked, "What sum, Terentius, would you consider
             "A million denarii, Sire."
             The Emperor grew pensive again. The general waited,
          his head down.
             Finally, the Emperor spoke:
             "Valiant Terentius, you are a great warrior and your
          prodigies of valour have earned you a lavish reward!
          I will give you wealth. Tomorrow at noon you will hear
             238-239   Stories about Giant   Numbers

             my decision."
              Terentius bowed and walked out.

             II. On next day at the hour appointed the general
             came to the Emperor's palace.
                "Greetings, brave Terentius!" the Emperor said.
                Terentius bowed his head humbly.
                "Sire, I came to hear your decision. You kindly
             promised to reward me."
                The Emperor answered: "It's not my intention that
             such a noble warrior like you should have some
             miserable reward for his heroic deeds. Listen to me.
             There are in my treasury 5 million copper brasses*.
             You shall go to the treasury and take one coin, then
Figure 228   you shall return here and place it at my feet. On the
             following day you shall again go to the treasury, take
             a coin worth 2 brasses and place it here near the first
             one. O n the third day you are to bring a coin worth
             4 brasses and on the fourth day bring a coin worth
             8 brasses, on the fifth, 16, etc., double the value of the
             previous coin. I will order appropriate coins be
             produced for you an<} while you have the strength, you
             will take them from my treasury. Nobody may help
             you, you must rely on your own power only. You will
             stop when you notice that cannot move a coin any
             more and then our deal will come to an end. All the
             coins that you will have managed to bring here will
             belong to you and you shall keep them as your
                Terentius listened eagerly to the Emperor's words.
             He visualized the multitude of coins, each one more
             than another, that he would bring out of the treasury.
                "I'm happy with your favour," he beamed. "Really
             generous is your reward!"

             III. Terentius started his daily visits to the treasury. It
             was located close to the Emperor's hall so the first trips
             with the coins cost Terentius very little effort.
                O n the first day he only brought 1 brass. This was
             a small coin 21 millimetres across and weighing
             5 grammes.
                His trips upto the sixth day were also very easy and
             he brought the coins double, fourfold, sixteen-fold, and
             thirty-two-fold the weight of the first.

                * A brass is a fifth of a denarius.
238-239   Stories about Giant   Numbers

             The seventh coin weighed 320 grammes and was
          8 1/2 centimetres across.*
             On the eighth day Terentius had to carry out a coin
          that was worth 128 units. It weighed 640 grammes and
          was about 10 1/2 centimetres wide.
             On the ninth day he carried into the Emperor's hall
          a coin corresponding to 256 unit coins. It was 13
          centimetres wide and weighed more than 1 1/4
             On the twelfth day the coin was almost 27
          centimetres across and 10 1/4 kilogrammes in weight.
             The Emperor who up until that day was very kind to
          the general now couldn't conceal his triumph. He saw
          that after 12 days only slightly more than 2,000 brass
          units had been brought.
             Further, on the thirteenth day the brave Terentius
          brought out a coin that was worth 4,096 units. It was
          34 centimetres wide and weighed 20 1/2 kilogrammes.
             On the fourteenth day Terentius had a heavy coin
          that was 42 centimetres across and weighed 41
             "Are you tired, my brave Terentius?" the Emperor
          could hardly help smiling.
             "No, Sire," the general responded grimly wiping his
             The fifteenth day came. This time Terentius's burden
          was really heavy. He trudged slowly to the Emperor
          carrying a huge coin corresponding to 16,384 unit
          coins. It was 53 centimetres wide and weighed 80
          kilogrammes, the weight of a tall warrior.
             On the sixteenth day the general staggered with the
          burden on his back. It was a coin equal to 32,768 units,
          its diameter being 67 centimetres and weighing 164
             The general was exhausted and gasping. The
          Emperor smiled...
             When Terentius came to the Emperor the next day,
          there was a roar of laughter. He could no longer carry
          his coin in his hands and rolled it in front of him. The
          coin was 84 centimetres and 328 kilogrammes, and
          corresponded to 65,536 unit coins.
             The eighteenth day was the last day of Terentius's
          enrichment, for his visits to the treasury and trips to
              * If a coin's volume is 64 times that of a normal one, then it is
          only four times wider and thicker, because 4 x 4 x 4 = 64. We
          should have this in mind when working out the sizes of further
238-239                Stories about Giant   Numbers

the Emperor's hall ended on that day. This time he had
to fetch a coin worth 131,072 unit coins. It was more
than a metre across and weighed 655 kilogrammes.
Using his spear as a lever Terentius rolled it into the
hall with a huge effort. The mammoth coin fell
thundering at the Emperor's feet.
  Terentius was completely worn out.
  "Can't do any more... Enough for me," he wispered.
  The Emperor could hardly conceal his pleasure at
the total triumph of his ruse. He ordered the treasurer
to compute the total of all the brasses brought into the
hall by Terentius.
   The treasurer reckoned quickly and said: "Sire,
thanks to your generosity the victorious warrior
Terentius has got a reward of 262,143 brasses."
   So the close-fisted Emperor gave the general about
1/20 part of the million of denarii Terentius had
                              *   *   *

  Let's check the treasurer's calculation and the weight
of the coins. Terentius brought out:
             Day          Coin             Weight
                          in brasses       in grammes
            1             1                5
            2             2               10
            3             4               20
            4             8               40
            5            16               80
            6            32              160
            7            64              320
            8           128              640
            9           256            1,280
           10           512            2,560
           11         1,024            5,120
           12         2,048           10,240
           13         4,096           20,480
           14         8,192           40,960
           15        16,384           81,920
           16        32,768          163,840
           17        65,536          327,680
           18       131,072          655,360

  The totals for these columns can be calculated easily
using the proper rule*, thus the second column totals

    * Each number in this column equals the sum of the previous
ones plus one. Therefore, when it's necessary to sum u p all the
numbers in the column, e. g. from 1 to 32,768, we need only find the
next number and subtract one, i.e. 32,768 x 2 - 1. The result is
238-239            Stories about Giant   Numbers

                   262,143. Terentius requested a million of denarii, i. e.
                   5 million brasses. Accordingly, he got
                      5,000,000 262,143 = 19 times
                   less than he requested.

                   Legend about Chess-Board
                   I. Chess is one of the world's most ancient games. It
                   has been in existence for centuries so it is no wonder
                   that it has given rise to many legends whose truth-
                   fulness cannot be checked because of the remotedness
                   of the events. One of these legends I want to relate.
                   You do not need to be able to play chess to understand
                   it, it is sufficient for you to know that it involves
                   a board divided into 64 cells (black and white
                      The play of chess was invented in India. When the
                   Indian king Sheram got to know about it he was
                   amazed at its ingeniousness and the infinite variety of
                   positions it afforded. Having learned that the play was
                   invented by one of his subjects, the king summoned
                   him in order to reward him personally for such
                   a stroke of brilliant insight.
                      The inventor, named Seta, came before the
                   sovereign's throne. He was a simply dressed scribe who
                   earned his living giving lessons to pupils.
Figure       229

16 — 9 7 5
238-239            Stories   about Giant   Numbers

   "I want to reward you properly, Seta, for the beaut-
iful play you invented," the king said.
   The sage bowed.
   "I'm rich enough to fulfil any of your desires," the
king went on to say. "Name a reward that would
satisfy you and you'll get it".
   There was a silence.
   "Don't be shy! What's your desire? I'll spare nothing
to meet your wish!"
   "Great is your kindness, oh sovereign. Give me some
time to sleep on it. Tomorrow, upon consideration, I'll
name you my wish."
   When the next day Seta came to the throne he
amazed the king by the unprecedented modesty of his
   Seta said: "Sovereign, order that one grain of wheat
be given to me for the first cell of the chess-board."
   "A simple wheat grain?" the king was shocked.
   "Yes, sovereign. For the second cell let there be two
grains, for the third four, for the fourth eight, for the
fifth 16, for the sixth 32..."
   "Enough!" the king was exasperated. "You'll get your
grains for all the 64 cells of the board according to
your wish: for each twice as much as for the previous
one. But let me tell you that your wish is unworthy of
my generosity. By asking for such a miserable reward
you show disrespect for my favour. Truly, as a teacher
you might give a better example of gratitude for the
kindness of your king. Go away! My servants will
bring you the bag of wheat."
   Seta smiled, left the hall and began to wait at the
palace gates.

II. At dinner the king remembered about the inventor
of chess and asked if the foolish Seta has collected his
miserable reward.
   The answer was: "Sovereign, your order is being
fulfilled. The court mathematicians are computing the
number of grains required."
   The king f r o w n e d - h e wasn't used to having his
orders fulfilled so slowly.
   At night, before going to bed the king Sheram again
inquired how long before had Seta left the palace with
his bag of wheat.
   "Sovereign, your mathematicians are working hard
and hope to finish their calculations before dawn."
   "Why so long?" the king was furious. "Tomorrow,
242-243   Stories   about Giant   Numbers

          before I wake up everything, down to the last grain,
          must be given to Seta. I never give my order twice!"
             First thing in the morning the king was told that the
          chief mathematician humbly asked to make an
          important report.
             The king ordered him in.
             Sheram said: "Before you bring out your business I'd
          like to know if Seta has at last received the miserable
          reward that he asked for."
             The old man responded: "It's exactly because of this
          that I dared to bother you at such an early hour. We've
          painstakingly worked out the number of grains that
          Seta wants to have. The number is so enormous..."
             "No matter how enormous it is", the king interrupted
          him arrogantly, "my granaries won't be depleted! The
          reward is promised and must be given out..."
             "It's beyond your power, oh sovereign, to fulfil his
          wish. There is not sufficient grain in all your barns to
          give Seta what he wants. And there is not enough in all
          the barns throughout the kingdom. You would not find
          that many grains in the entire space of the earth. And if
          you wish to give out the promised reward by all means,
          then order all the kingdoms on earth to be turned into
          arable fields, order all the seas and oceans dried up,
          and order the ice and snowy wastes that cover the far
          northern lands melted. Should all the land be sown
          with wheat and should the entire yield of these fields be
          given to Seta, then he'd receive his reward."
             The king attended to the words of the elder with
             "What is this prodigious number?"
             "18,446,744,073,709,551,615, oh sovereign!"

          III. Such was the legend. There is now no way of
          knowing if it's true, but that the reward is expressed by
          this number you could verify by some patient
          calculations. Starting with unity you'll have to add up
          number 1,2,4,8, etc. The result of the 63th doubling will
          be what the inventor should receive for the 64th cell of
          the board.
             If you use the rule explained at the end of the
          previous problem, you can easily obtain the number of
          grains to be received by the inventor (we double the
          last number and subtract one). Hence the calculation
          comes down to multiplying together 64 twos:
          2 x 2 x 2 x 2 x 2 , etc.,-64 times.
238-239              Stories   about   Giant   Numbers

   To facilitate computation divide the 64 multipliers
into six groups with 10 twos in each and one last group
with four twos. It's easy to see that the product of 10
twos is 1,024, and of four twos, 16. The desired result is
thus 1,024 x 1,024 x 1,024 x 1,024 x 1,024 x 1,024 x 16.
   Multiplying 1,024 x 1,024 gives 1,048,576.
   It now remains to find 1,048,576 x 1,048,576 x
 x 1,048,576 x 16, subtract one from the result to arrive
at      the       sought-for  number       of      grains:
   If you want to imagine the enormousness of this
numerical giant just estimate the size of a barn that
would be required to house this amount of grain. It's
known that a cubic metre of wheat contains about
15,000,000 grains. Consequently, the reward of the
inventor       of     chess  would      occupy      about
12,000,000,000,000 cubic metres, or 12,000 cubic
kilometres. If the barn were 4 metres high and 10
metres wide its length would be 300,000,000 kilometres,
twice the distance to the Sun!
   The Indian king could never grant such a reward.
Had he been good at maths, he could have freed him-
self of the debt. He should have suggested to Seta to
count off the grains he wanted himself.
   In fact, if Seta kept on counting day in day out he
would have counted only 86,400 grains in the first 24
hours. A million would have required no less than 10
days of continual reckoning and thus to process 1 cubic
metre of wheat would have required about half a year.
In a ten year's time he would have handled about 20
cubic metres. You see that even if Seta had devoted
a lifetime to his counting, he would still have only
obtained a miserable fraction of the reward he desired.

Prolific   Multiplication
A ripe poppy head is full of tiny seeds, each of which
can give rise to a new plant. How many poppy plants
shall we have, if all the seeds germinate? To begin with
we should know how many seeds there are in a head.
A boring business, but if you summon up all your
patience you'll find that one head contains about 3,000
  What follows from this? If there is enough space
around our poppy plant with adequate soil, each seed
will produce a shoot with the result that the following
244-245   238-239   Stories about Giant   Numbers

          summer 3,000 poppies will grow. A whole poppy field
          from just one head.
            Let's see what will happen next. Each of the 3,000
          plants will produce no less than one head (more often
          several heads), with 3,000 seeds each. Having
          germinated, the seeds of each head will give 3,000 new
          plants, and hence during the following year we are
          going to have
          3,000 x 3,000 = 9,000,000 plants.
            Calculation gives that in the third year the offspring
          of our initial head will already reach
          9,000,000 x 3,000 = 27,000,000,000.
            In the fourth year there will be
          27,000,000,000 x 3,000 = 81,000,000,000,000 offspring.
            In the fifth year our poppies will engulf the earth,
          because they'll reach the number
          81,000,000,000,000 x 3,000 = 243,000,000,000,000,000.
             But the surface area of all the land, i.e. all the
          continents and islands of the earth, amount to
          135,000,000 square kilometres, or 135,000,000,000,000
          square m e t r e s - a b o u t 2,000 times less than the number
          of the poppy plants grown.
             You see thus that if all the poppy-seeds from one
          head germinated, the offspring of one plant could
          engulf the earth in five years so that there were about
          2,000 plants of each square metre of land. Such
          a numerical giant lives in a tiny poppy seed!
             A similar calculation made for a plant other than the
          poppy, one which yields less seeds, would lead to the
          same result with the only distinction that its offspring
          would cover the lands of the earth in a longer period
          than five years. Take a dandelion, say, which gives
          about 100 seeds annually. Should all of them germinate,
          we would have:
                    Year                      N u m b e r of plants
                    1                                                 1
                    2                                             100
                    3                                          10,000
                    4                                       1,000,000
                    5                                     100,000,000
                    6                                  10,000,000,000
                    7                               1,000,000,000,000
                    8                             100,000,000,000,000
                    9                          10,000,000,000,000,000
238-239               Stories about Giant   Numbers

   This is 70 times more than the square metres of land
available on the globe.
   In consequence, the whole Earth would be covered
by dandelions in the ninth year with about 70 plants on
each square metre.
   Why then don't we observe in reality these
tremendous multiplications? Because the overwhelming
majority of seeds die without producing any new
plants, they either fail to hit a suitable patch of soil and
don't germinate at all, or having begun to germinate
are suppressed by other plants, or are eaten by animals.
If there were no massive destruction of seeds and
shoots, any plant would engulf our planet in a short
   This is true not only of plants but of animals, too. If
it were not for death, the offspring of just one couple of
any animal would sooner or later populate all the land
available. Swarms of locust covering huge stretches of
land may give some idea of what might happen on
earth if death didn't hinder the multiplication of living
things. In two decades or so the continents would be
covered with impenetrable forests and steppes inhabited
by incountable animals struggling for their place under
the sun. The oceans would be filled to the brim with
fish so that any shipping would be impossible. And the
air would not be transparent because of the mists of
birds and insects...
   Before we leave the subject, we'll consider several
real-life examples of uncannily prolific animals placed
in favourable conditions.
   At one time America was free of sparrows. The bird
that is so common in Europe was deliberately brought
to the United States to have it exterminate the
destructive insects. The sparrow is known to eat in
quantity voracious caterpillars and other garden and
forest pests. The sparrows liked their new environment,
since there were no birds of prey eating them, and so
they began to multiply rapidly. The number of insects
began to drop markedly and before long the sparrows,
for want of animal food, switched to vegetable food and
went about destroying crops*. The Americans were
even forced to initiate a sparrow control effort which
appeared to be so expensive that a law was passed
forbidding the import to America of any animals.

    * In the Hawaii they even completely superseded small endemic
          Stories   about Giant   Numbers

             A further example. There were no rabbits in Australia
          when the continent was colonized by the Europeans.
          The rabbits were brought to Australia in the late 18th
          century and as there were no carnivores that might be
          their enemies they began to multiply at a terrifically
          fast rate. Hordes of rabbits soon inundated Australia
          inflicting enormous damage to agriculture. They
          became a plague of the country and their eradication
          required great expense and effort. Later the same
          situation with rabbits occurred in California.
             A third instructive story comes from Jamaica. The
          island was suffering from an abundance of poisonous
          snakes. To get rid of them it was decided to introduce
          the secretary-bird, an inveterate killer of poisonous
          snakes. The number of snakes soon dropped all right,
          but instead the island got to be infested with the rats
          that earlier were controlled by the snakes. The rats
          wrought dreadful havoc amongst the sugar cane fields
          and posed an urgent problem. It's known that an
          enemy of the rats in the Indian mongoose, and so it was
          decided to bring four pairs of these animals to the
          island and allow them multiply freely. The mongooses
          adapted perfectly to their new land and in a short
          period of time inhabited the island. In less than
          a decade they had almost wiped out the rats. But alas,
          having destroyed the rats, the mongooses began to
          consume whatever came their way and turned into
          omnivores. They started killing puppies, goat-kids,
          piglets, poultry. And when they had multiplied still fur-
          ther they set about devastating orchards, fields and
          plantations. So the inhabitants of the island were
          compelled to start combating their previous allies, but
          with limited success.

          Free Dinner
          Ten young people decided to celebrate leaving school
          by a dinner at a restaurant. When all had gathered they
          started arguing as to how they were to sit at the table.
          Some suggested that they sit in alphabetic order, others,
          by age, yet others, by their academic record, or even by
          their height.
             The argument dragged on, but nobody sat down at
          the table.
             It was the waiter who made it up between them. He
             "My young friends, you'd better stop arguing, sit at
238-239            Stories   about   Giant   Numbers

the table arbitrarily and listen to me."
   The ten set anyhow and the waiter continued:
   "Let somebody record the order in which you are
sitting now. Tomorrow come here again to dine and sit
in another order. The day after tomorrow you sit in
a new order, and so on, until you have tried out all the
arrangements possible. When you come to sit in exactly
the same order as you are sitting now, then upon my
word, I'll start to treat you to the finest dinners without
   The party liked the suggestion. It was decided to
come every night and try all the ways of sitting at the
table in order to enjoy the free dinner as soon as
   They didn't live to see it, however. And not because
the waiter didn't keep his word, but because the
number of arrangements was too great. Specifically, it is
3,628,800. You can see this number of days equals to
almost 10,000 years!
   It might seem unlikely to you that as few as 10
people might be arranged in such an enormous number
of various ways but we can check it.
   To begin with, we must learn how to find the
number of permutations possible. To make our life
easier we'll begin with a small number of objects, say
three. Let's label them A, B, and C.
   We would like to know in how many ways it's
possible to permute them. We argue as follows: if for
the moment we put B aside, the two remaining objects
may be arranged in two ways. We will now attach B to
each of the two pairs. We may place it in each of three
   (1) B behind the pair;
   (2) B in front of the pair;
   (3) B between the members of the pair.
   Clearly, there are no other positions for B besides
these three. But as we have two pairs, AB and BA, then
there are 2 x 3 = 6 ways of arranging the objects.
   Further, we'll repeat the argument for four objects.
Let there be four objects A, B, C, and D. We'll again
put aside one object and make all the possible
permutations with the remaining three. We know
already that there are six of these. In how many ways
may we attach D to each of the six arrangements of
three? Obviously, we may place it as follows;
   (1) D behind the triple;
   (2) D in front of the triple;
238-239   Stories   about Giant   Numbers

             (3) D between the first and second objects;
             (4) D between the second and third objects.
             We thus get 6 x 4 = 24 permutations, and since 6 =
           = 2 x 3 , and 2 = 1 x 2 ,      then the number of
          permutations may be represented as the product 1 x
           x 2 x 3 x 4 = 24.
             Reasoning along the same lines we'll find that for five
          objects, too, the number of permutations is 1 x 2 x 3 x
           x 4 x 5 = 120.
             For six objects: 1 x 2 x 3 x 4 x 5 x 6 = 720, and so
             Return now to the case of the 10 diners. The number
          of permutations possible here is obtainable if we take
          the trouble of multiplying together I x 2 x 3 x 4 x 5 x
           x 6 x 7 x 8 x 9 x l 0 . Tnis will in fact give the
          above-mentioned 3,628,800.
             The calculation would be more complex if among the
          10 diners there were five girls who wanted to alternate
          with the boys. Although the number of the possible
          permutations is far less in this case, it's somewhat more
          difficult to work it out.
             Let one boy seat at the table somewhere. The
          remaining four may only be seated in alternate chairs
          (leaving the vacant places for the girls) in 1 x 2 x 3 x
           x 4 = 24 various ways. Since the total number of
          chairs is 10, the first boy may be seated in 10 ways,
          hence the number of all possible arrangements for the
          boys is 10 x 24 = 240.
             What is the number of ways in which the five girls
          may occupy the vacant chairs between the boys?
          Clearly, 1 x 2 x 3 x 4 x 5 = 120. Combining each of the
          240 positions for the boys with each of the 120
          positions of the girls we obtain the number of all the
          possible arrangements: 240 x 120 = 28,800.
             The above number is smaller by far than the
          previous one though it would require almost 79 years
          to work through them all. Should the young guests of
          the restaurant live to be 100, they could get the free
          dinner, if not from the waiter himself, then from his
Tricks with Numbers

Out of Seven Digits
Write the seven digits from 1 to 7 one after the other:
1 2 3 4 5 6 7.
  It's easy to connect them by the plus and minus signs
to obtain 40, e.g.
12 + 34 - 5 + 6 - 7 = 40.
 Try and find another combination of these digits that
would yield 55.

Nine Digits
Now write out the nine digits: 1 2 3 4 5 6 7 8 9.
  You can as above arrive at 100 by inserting a plus or
minus six times and get 100 thus:
12 + 3 - 4 + 5 + 67 + 8 + 9 = 100.
  If you want to use only four plus or minus signs, you
proceed thus:
123 + 4 - 5 + 6 7 - 8 9 = 100.
  Now try and obtain 100 using only three plus or
minus signs. It's much more difficult but possible.

With Ten Digits
Obtain 100 using all ten digits.
  In how many ways can you do it? There are no less
than four different ways.


Obtain unity using all ten digits.

With Five Twos
We only have five twos and all the basic mathematical
operation signs at our disposal. Use them to obtain the
following numbers: 15, 11, 12, 321.

Once More with Five Twos
Is it possible to obtain 28 using five twos?
250-251   Tricks with   Numbers

          With Four Twos
          The problem is more involved. Use four twos to arrive
          at 111. Is that possible?

          With Five Threes
          To be sure, with the help of five threes and the
          mathematical operation signs we can represent 100 as

          33 x 3 + y = 100.

            But can you write 10 with five threes?

          The Number 37

          Repeat the above problem to obtain 37.

          In Four Ways
          Represent 100 in four various ways with five identical

          With Four Threes
          The number 12 can be very easily expressed with four
          12 = 3 + 3 + 3 + 3.
            It's more of a problem to obtain 15 and 18 using
          four threes:
          15 = ( 3 + 3) + ( 3 x 3);
          18 = (3 x 3 ) + (3 x 3).
           And if you were required to arrive at 5 in the same
          way, you might be not very quick to twig that 5 =

             Now think of the ways to get the numbers 1, 2, 3, 4,
          6, 7, 8, 9, 10.

          With Four Fours
          If you have done the previous problem and want some
          more in the same vein, try to arrive at all the numbers
256-253            Tricks with   Numbers

from 1 to 10 with fours. This is no more difficult than
getting the same numbers with the threes.

With Four Fives
Obtain 16 using four fives.

With Five Nines
Can you provide at least two ways of getting 10 with
the help of five nines?

It's very easy to obtain 24 with three eights: 8 + 8 + 8.
Could you do this using other sets of three identical
digits? The problem has several solutions.

The number 30 can easily be expressed with three fives:
5 x 5 + 5. It's more difficult to do this with other sets of
identical digits. Try it, you'll may be able to find several

One Thousand
Could you obtain 1,000 with the aid of eight identical

Get Twenty
The following are three numbers written one below the
Try and cross out six digits so that the sum of the
remaining numbers be 20.

Cross out Nine Digits
The following columns of five figures each contain 15
odd digits:
252-253   Tricks with   Numbers

          1 1   1
          3 3   3
          5 5   5
          9 9   9
            The problem is to cross out nine digits so that the
          numbers thus obtained add up to 1,111.

          In a Mirror
          The number corresponding to a year of the last century
          is increased 41/2 times if viewed in a mirror. Which
          year is it?
          Which     Year?
          In this century, is there a year such that the number
          expressing it doesn't change if viewed "upside down"?

          Which     Numbers?
          Which two integers, if multiplied together, give 7?
          Don't forget that both numbers should be integers,
          therefore answers like 3 1 / 2 x 2 or 2 1 / 3 x 3 won't do.

          Add and       Multiply
          Which two integers, if added up, give more than if
          multiplied together?

          The Same
          Which two integers, if multiplied together, give the
          same as if added up?
          Even Prime        Numbers
          You must know that prime numbers are those that are
          divisible without remainder by themselves only or by
          unity. Other numbers are called composite.
            What do you think: are all the even numbers
          composite? Are there any even prime numbers?

          Three     Numbers
          Which three integers, if multiplied together, give the
          same as if added up?
258-253              Tricks with   Numbers

Addition and      Multiplication
You've undoubtedly noticed the curious feature of the
2 + 2 = 4,
2 x 2 = 4.
   This is the only case where the sum and product of
two integers (equal to each other at that) are equal.
   You maybe are unaware that there are dissimilar
numbers showing the same property. Think of examples
of such numbers. So that you don't believe that the
search would be in vain I assure you that there are
many such number pairs, though none of them are

Multiplication    and Division
Which two integers yield the same result whether the
larger of them is divided by the other or they are
multiplied together?

The Two-Digit      Number
There is a two-digit number such that if it is divided by
the sum of its digits the answer is also the sum of the
digits. Find the number.

Ten Times More
The numbers 12 and 60 have a fascinating property: if
we multiply them together, we get exactly 10 times
more than if we add them up:
12 x 60 = 720, 12 + 60 = 72.
Try and find another pair like this. Maybe you can find
several pairs with the same property.

Two Digits
What is the smallest positive integer that you could
write with two digits?

The Largest      Number
What is the largest number that you can write with
four ones?
254-255   7ticks     with   Numbers

          Unusual Fractions
          Consider the fraction 6729/13,458. All the digits (save
          for 0) are used in it once. As is easily seen, the fraction
          is 1/2.
             Use the nine digits to obtain the following fractions:
                 1 1 1 J_
                 J' J' J' 7' J_ J_'
                             8~' 9

          What Was the            Multiplier?
          A schoolboy carried out a multiplication, then rubbed
          most of his figures from the blackboard so that only
          the first line of the figures and two digits in the last line
          survived. As to the other figures, only the following
          traces remained:

                 * * * *

               * * * #


               Could you restore the multiplier?

          Missing Figures
          In this multiplication case more than half the figures
          are replaced by asterisks:


          Can you restore the missing figures?

          What Numbers?
          A further problem of the same sort:
260-253            Tricks with   Numbers




Strange Multiplication   Cases
Consider the following case of the multiplication of two
48 x 159 = 7,632.
It's remarkable in that each of the nine digits is
involved once here. Can you think of any other
examples? If so, how many of them are there?

Mysterious   Division
What is given below is nothing but an example of
a long-division sum where all the digits are replaced by
               . . 7.

Not one digit in either the dividend or the divisor is
known. It's only known that the last but one digit in the
quotient is 7. Determine the result of the division.
  The problem has only one answer.
256-253        Tricks   with    Numbers

               Another Division Problem
               Restore the missing figures in the division below:
               325) ***


Figure   230              *5*

               Division by 11
               Write out a nine-digit number containing no repeated
               digits (all the digits are different), that divides by 11
               without remainder.
                   What is the largest such number?
                   What is the smallest such number?

               Triangle of Figures
Figure   231
               Within the circles of the triangle of Fig. 230 arrange all
               the nine digits so that the sum of the digits on each
               side be 20.

               Another         Triangle
               Repeat the previous problem so that each side adds up
               to 17.

               Eight-Pointed        Star
Figure   232   Into the circles of the figure of Fig. 231 insert one of
               the numbers from 1 to 16 so that the sum of the
               numbers on the side of each square be 34 and the sum
               of the numbers at the corners of each square be 34, too.

               Magic Star
               The six-pointed star shown in Fig. 232 is "magic"
               because all the six lines of numbers have the same sum:
               4 + 6 + 7 + 9 = 26, 1 1 + 6 + 8 + 1 = 26,
               4 + 8 + 12 + 2 = 26, 1 1 + 7 + 5 + 3 = 26,
               9 + 5 + 10 + 2 = 26,      1 + 1 2 + 1 0 + 3 = 26.
             Tricks   with   Numbers

               However the numbers at the points of the star add
             up to another number:
Figure 233
             4 + 1 1 + 9 + 3 + 2 + 1 = 30.
               Couldn't you improve the star so that the numbers at
             the points also gave the same sum (26)?

             Wheel of Figures
             The digits from 1 to 9 should be so arranged in the
             circles of the wheel of Fig. 233 that one digit is at the
             centre and the others elsewhere about the wheel so that
             the three figures in each line add up to 15.

             It's required to arrange the numbers from 1 to 13 in the
Figure 234   cells of the trident shown in Fig. 234 so that the sums
             of the figures in each of the three columns (I, II, and
             III) and in the line (IV) are the same.
258-283                        Answers

                               Out of Seven Digits
There are three solutions:
                               123 + 4 - 5 - 6 7 = 55;
                               1 - 2 - 3 - 4 + 56 + 7 = 55;
                               12 - 3 + 45 - 6 + 7 = 55.

                               Nine Digits
                               123 - 45 - 67 + 89 = 100.
This is the only solution. It's impossible to arrive at the same result by using the plus
and minus signs less than three times.

                               With Ten Digits
The following are the four solutions:

                               70 + 2 4 ^ - + 5 \ = 1 0 0 ;
                                        18       6
                                  27     3
                               80 — + 1 9 - = 100;
                                  54     6

                               84 + 9 ^ - + 3 ^ - = 100;
                                      j       bU
                                  1      38
                               50 — + 49 — = 100.
                                  2      76

                               U nity
Represent unity as the sum of two fractions:
                                148     35^_1

Those knowing more advanced mathematics may also give other answers:
                                123,456,789°; 234,5679 " 8 ~ \ etc.,
since any number to the zeroth power is unity.

                               With Five Twos
                               Write 15 as:

                               (2 + 2)2 — y = 15;                 - ^ - + 2 x 2 = 15;

                               (2 x 2)2 — y = 15;                  ~   +

                               2(2 +   2)_|   = 1 5   .           - ^ - + 2 + 2 = 15.

And 11 as:
                               — + 2-2=11.

Now the number 12,321. At first sight, it would seem impossible to write this five-digit
number with five similar figures. The problem is manageable, however. Here is the
                                 222 Y
                                          = ill2 = i n        x    111 = 12,321.

                               Once More with Five Twos

                               22 + 2 + 2 + 2 = 28.

                               With Four Twos

                               ^==- = 111.

                               With Five Threes
The solution is:
                               33    3
                              ^ - 4 = 1 0 .
                                3    3
    It's worth mentioning that the problem would have had exactly the same solution
if we had had to express 10 with five ones, five fours, five sevens, five nines, or, in
general, with any five identical digits. In fact:
                               11        1    22          2   44       4   99      9
                                           =r               —            —               PtC
                                1        1     2          2    4       4    9           9 ' '
260-261                      Answers

Also, there are other solutions to the problem:
                             33  3    _
                             — + — = 10.
                              3  3

                             The Number 37
There are two solutions:
                             33 + 3 + y = 37;


                             In Four Ways
We can use ones, threes and (most conveniently) fives:
                                    111 - 11 = 100;

                                         33 x 3 + y = 100;

                              5 x 5 x 5 - 5 x 5 = 100;
                             (5 + 5 + 5 + 5) x 5 = 100.

                              With Four Threes
                              1           (there are also other ways);

                              3 =   3+3+3
                                    3x3 + 3
                              4 =

                              6 = (3 + 3 ) x j .
                             X n.swers

  We've given the solutions through six only. Work out the remaining ones for
yourselves. The above solutions, too, may be represented with other combinations of
                             With Four Fours

                                    44      4+ 4                         4x4
                             1 =                                               etc.;
                                   4 4 ' ° 4 ^ 4 ' °r                    474
                                    4               4         4 x4
                             2              +
                                 =4                 4' °r     4T4    ;

                                    4+4+4                        4x4-4
                             3=        :  , or                     -   ;

                             4 = 4 + 4 x (4-4);
                                 _ 4x 4+ 4
                             5        ~    •

                                        4       +   4
                                 *           ,
                                 6 = — - — + 4;

                                            4     44
                                 7 = 4 + 4 — —, or - - 4 ;

                                 8 = 4 + 4 + 4 - 4 , or                  4x4-4-4;
                                                        4 '
                                 10 =

                             With Four Fives
There is only one way:
                             — + 5 = 16.

                             With Five Nines
The two ways are as follows:
262-283                      A nswers

                             99         9
                              9         9            '

Those knowing more mathematics may add several other solutions, e.g.

                              9 + | - V = 10, or 9 + 99 9 ~ 9 = 10.

The two solutions are:

                             22 + 2 = 24;                3 3 - 3 = 24.

The three solutions are
                             6 x 6 - 6 = 30;                3 3 + 3 = 30;   33 - 3 = 30.

                             One Thousand

                             888 + 88 + 8 + 8 + 8 = 1,000.

                             Get      Twenty
The crossed out digits are replaced by zeros:
because 11 + 9 = 20.

                             Cross Out Nine Digits
The problem permits of several solutions. We                furnish four:
                              100 111 011                    101
                             000 030 330                     303
                             005 000 000                    000
                             007 070 770                    707
                             999 900 000                    000

                             1111 1111 1111 1111
                                A nswers

                                In a Mirror
The only figures that are not inverted in a mirror are 1, 0, and 8. Accordingly, the
year we seek can only contain these figures. Besides, we know that the year is in the
19th century, hence the first two figures are 18.
   It's easily seen now that the year is 1818 because in a mirror we obtain 8181, which
is 41/2 times more than 1818. In fact,
                                1818 x 41/2 = 8181.
     The problem has no other solutions.

                                Which      Year?
In the 20th century there is only one such year, viz. 1961.

                                Which      Numbers?
The answer is simple: 1 and 7. There are no other numbers.

                               Add and       Multiply
There are many such numbers, e.g.
                                 3x1=3;              3 + 1=4;
                                1 0 x 1 = 10; 1 0 + 1 = 11.
  In general, any pair of integers of which one is unity will work.
  This is because adding one increases a number and multiplying by one does not
change it.

                                The Same
The numbers are 2 and 2. There are no other integers.

                               Even Prime          Numbers
There is one even prime n u m b e r - 2 . It only divides by itself and 1.

                               Three       Numbers
Multiplying 1, 2, and 3 gives the same as adding them up:
                               1 + 2 + 3 = 6,        1 x 2 x 3 = 6.
264-265                        Answers

                               Addition     and         Multiplication
There are a lot of such pairs. Examples are:
                                    3 + 1 1 / 2 = 4 1/2,               3 x 1 1/2 = 4 1/2,
                                    5 + 1 1 / 4 = 6 1/4,               5   X   1 1/4 = 6 1/4,
                                    9 + 1 1/8 = 10 1/8,                9   X   1 1/8 = 10 1/8,
                                11 + 1.1 = 12.1,                      11   X   1.1 = 12.1
                                21 + 1 1/20 = 22 1/20,                21   X   1 1/20 = 22 1/20,
                               101 + 1.01 = 102.01,                  101   X   1.01 = 102.01, etc.
                               Multiplication          and Division
There are many correct number pairs. For example,
                                2 4-1=2,                2x1=2,
                                7 - ^ 1 = 7,            7 x 1 = 7,
                               43      1 = 43,         43 x 1 = 43.

                               The Two-Digit             Number
The number we seek should clearly be a square. As among the two-digit numbers
there are only six squares, then by trial-and-error method we readily find the unique
solution, namely 81:

                                         = 8 + 1
                               8TT                 -

                               Ten Times         More
The following are the four other pairs of such numbers: 11 and 110; 14 and 35; 15
and 30; 20 and 20.
In fact,
                               11x110=1,210;                 11 + 110 = 121;
                               14 x      35 = 490;           14+      35 = 49;
                               15 x      30 = 450;           15 +     30 = 45;
                               20 x      20 = 400 ;         20+       20 = 40.
   The problem has no other solutions. Searching for the solutions by trial and error is
tiresome and a knowledge of the ABC of algebra would make the process easier and
enable us not only to find all the solutions, but also to make sure that the problem
doesn't have more than five solutions.
                               A nswers

     f-                       Two Digits
Many may believe that the number is 10. No, it's                         1, expressed as follows:
                              1 2 3 4                                     9
                                                    etc       up   to
                              7' 7' 7' T'                 "              7"
  Those who know some more mathematics may add to these answers a number of
                               1°, 2°, 3°, 4°, etc., up to 9°,
because any number to the zeroth power is unity*.

                               The Largest       Number
The commonest answer is 1,111. But the number is far from being the largest, l l 1 1 is
much more, 250,000,000 times more.

                              Unusual        Fractions
There are several solutions. One of them is
                               1       5,823       1           3,942
                               3      17,469       4          15,768

                               1       2,697        1          2,943
                               5      13,485        6         17,658 '

                               1       2,394        1          3,187          1    6,381
                               7      16,758        8         25,496 '        9   57,429 '

There are many versions, especially the ones for 1/8 of which there are more than 40.

                              What Was the          Multiplier?
We argue as follows. The figure 6 is the result of the addition of two figures, one of
which may be either 0 or 5. But, if the lower one is 0, then the upper is 6. But, may
the upper figure be 6? Testing shows that whatever the second figure of the multiplier,
6 cannot be in the last but one place of the first partial product. Accordingly, the
lower figure of the last but one column must be 5 and above it 1.

                                   * But the solutions — or 0° would be wrong because they are
266-261                          Answers

  We can now easily restore some of the missing figures:


   The last figure in the multiplier must be more than 4, otherwise the first partial
product will not consist of four figures. It cannot be 5 as we won't then get 1 in the
last but one place. Let's try 6 and this works out all right. We have:

                                     1 410

  Reasoning further along the same lines we find that              the multiplier    is 96.

                                Missing          Figures
The missing figures are restored one after another if we use the following argument.
  For convenience we assign numbers to the lines
                                     *i*          I
                                     3*2        II
                                   *3*          III
                                 3*2*           IV
                              + *2*5             V
                                1*8*30          VI
  It's easily seen that the last asterisk in line III is 0, which follows from the fact that
0 is at the end of line VI.
  Now we determine the value of the last asterisk in line I. This figure must give
a number ending in zero when multiplied by 2, and a number ending in 5 (line V)
when multiplied by 3. There is only one such figure-5.
  We can now guess the meaning of the asterisk in line II. It's 8 because only when
multiplied by 8 the number 15 gives a result ending in 20 (IV).
  Eventually, the meaning of the first asterisk in line I becomes clear: it is 4, because
only 4, when multiplied by 8, gives a result beginning with 3 (line IV).
  The remaining figures now present no problem. Suffice it to multiply together the
                             A nswers

numbers of the first two lines that have                now       been   completely     determined.
  We'll end up with the multiplication:
                             What        Numbers?
Arguing as above we uncover the meaning of the asterisks in this case, too. We get
                             +     1300

                            Strange Multiplication            Cases
The patient reader can find the following nine cases where the                        multiplication
calculations meet the question's demands. They are:
                              12 x 483 = 5,796, 48 x 159 = 7,632,
                              42 x 138 = 5,796, 28 x 157 = 4,396,
                              18 x 297 = 5,346, 4 x 1738 = 6,952,
                              27 x 198 = 5,346, 4 x 1963 = 7,852.
                              39 x 186 = 7,254,
                             Mysterious           Division
For convenience, we'll number the lines in the arrangement thus:
Looking at line II we conclude that the second figure of the quotient is 0 as it was
necessary to borrow two figures from the dividend. Denote the divisor by x. Lines IV
and V indicate that Ix (the product of the last but one digit in the quotient and the
divisor), when it was subtracted from a number not larger than 999, gave a difference
not less than 100. Clearly, Ix cannot exceed 999 — 100, i. e. 899, hence x is not larger
than 128. Further, we see that the number in line III is more than 900, otherwise, it
wouldn't give a two-digit difference when subtracted from a four-digit number. Thus
the third digit of the quotient should be 900 —128, i.e. more than 7.03. Accordingly, it
is either 8 or 9. The numbers in I and VII are four digits long hence the third digit in
the quotient is 8 and the extreme left and right digits are 9.
   This actually completes the problem as the desired result (the quotient) has now
been found. It is 90,879.
   It's not necessary to go on with the argument to find the dividend and divisor, as
we wanted the quotient only. The problem doesn't require us to decipher the whole of
the arrangement. Besides, there are 11 pairs of numbers that satisfy the given
arrangement of points and give 7 in the fourth place of the quotient, viz.
                                 10,360,206 114"^
                                 10,451,085 115
                                 10,541,964 116
                                 10,632,843 117
                                 10,723,722- 118
                                 10,814,601 119 = 90,879.
                                 10,905,480 120
                                 10,996,359 121
                                 11,087,238 122
                                 11,178,117 123
                                 11,268,996 124

                              Another Division Problem
The answer is:

                               325)     52650





                                 A nswers

                                 Division by 11
To solve the problem requires the knowledge of the criterion of divisibility by 11.
A number is divisible by 11 if the difference between the sum of the even digits and
the sum of the odd digits either is divisible by 11 or is zero.
  By way of example, test the number 23,658,904.
  The sum of the digits in the even places is
                              3 + 5 + 9 + 4 = 21
and the sum of the digits in the odd places is
                                 2 + 6 + 8 + 0 = 16.
     Their difference (we subtract from the largest) is
                                 21 - 16 = 5.
  The difference (5) doesn't divide by 11 nor is it zero, hence the number under
consideration doesn't divide by 11.
  Take another number-7,344,535:
                                     3 + 4 + 3 = 10;
                                 7 + 4 + 5 + 5 = 21;
                                      21 - 1 0 = 11.
  As 11 divides by 11, the number in question is a multiple of 11.
  Now we can easily work out the order in which we should write the nine digits so
as to arrive at a number that is a multiple of 11 and meets the conditions of the
  Yet another example is 352,049,786. Let's test it:
                                 3 + 2 + 4 + 7 + 6 = 22;
                                    5 + 0 + 9 + 8 = 22.
     The difference is 22 — 22 = 0, hence the number divides by 11.
     Of these numbers the largest is 987,652,413, and the smallest is 102,347,586.

                                 Triangle of Figures
The solution is shown in Fig. 235. The figures in the middle of each line can be
interchanged to obtain further solutions.

Figure   235
270-283                         A nswers

                                Another    Triangle
Again, the solution is given in Fig. 236. Also, the figures in the middle of each line can
be interchanged to obtain further solutions.

Figure   236

The solution is in Fig. 237.

Figure   237

                               Magic Star
To make our life easier we'll start off with the following considerations.
   The numbers at the points must add up to 26, but the sum of all the numbers in the
star is 78. Accordingly, the numbers on the internal hexagon is 78 - 26 = 52.
   Now let's look at one of the large triangles. On each side we have 26, and for all the
three we get 26 x 3 = 78, with each number at the vertex entering twice. But as the
sum of the three internal pairs (i.e. of the internal hexagon) is known to be 52, then
the double sum of the numbers at the vertices of each triangle is 78 — 52 = 26, the
single sum is thus 13.

Figure   238

  The field of search has now been narrowed markedly. We know, for instance, that
neither 12 nor 11 can be at the star points (why?). Hence the tests might be begun
with 10. In that case we immediately determine the two numbers that must occupy the
remaining vertices of the triangle: 1 and 2.
  Moving on after this manner we eventually arrive at the desired arrangement. It's
shown in Fig. 238.

                              Wheel of Figures
The solution is given in Fig. 239.

Figure 239

The following is the desired arrangement (Fig. 240). The sum of the numbers in each
of the four lines is 25.

Figure 240

             Merry Arithmetic
             Simple   Multiplication
             If you don't remember the multiplication table properly
             and have difficulty in multiplying by 9, then your own
             fingers might be of help.
                Place both hands on a table, your 10 fingers will be
             your computer.
                Suppose you want to multiply 4 by 9.
                Your fourth finger gives the answer : on the left of it
             there are three fingers, on the right, six. So you read:
             36, hence 4 x 9 = 36.
                Further examples: how much is 7 x 9 ?
                Your seventh finger has six fingers on its left and
             three on its right. The answer: 63.
                What is 9 x 9? On the left of the ninth finger there
             are eight fingers, on the right, one. The answer is 81.
                This living computer will remind you, for example,
             that 6 x 9 is 54, not 56.

             Cats and Mats
             Once some cats
             found some mats.
             But if each mat
             had but one cat
             there's be a cat
             without a mat.
             Should each mat
             now have two cats
             there'd be a mat
             without a cat.
             How many cats
Figure 241   and how many mats?

             Sisters and Brothers
             I have an equal number of sisters and brothers. But my
             sister has two times more brothers than sisters. How
             many are we?

             How Many      Children?
             I have six sons. Each son has a sister. How many
             children have I?
             Merry   Arithmetic

 vP          Breakfast
             Two fathers and two sons breakfasted on three eggs,
             each having a whole egg. How do you account for it?

             Three Quarters of a Man
             A team leader was asked how many people there were
             in his team. He answered in a rather involved way:
             "Not many: three quarters of us plus three quarters of
             a man, that's all."
               Could you compute the number of people in his

             How Old Are          They?
             "Gran'pa, please tell me how old your son is?"
             "His is as many weeks old as my grandson is days old."
             "And your grandson, how old is he?"
             "His age is as many months old, as I am years old."
             "How old are you, then?"
             "Together the three of us are 100 years old. Now guess
             how old each of us is."

Figure 242   Who is Older?
             In two years my       boy    will be twice as old as he was two
             years ago. And        my     girl in three years will be three
             times as old as       she    was three years ago.
               Who is older,        my    boy or my girl?

             The Age of the Son
             Now my son is a third my age. But five years ago he
             was a quarter my age.
               How old is he?

             His Age
             A witty person was asked about his age. The answer
             was: "If you take thrice my age in three years and
             subtract thrice my age three years back, then you'll
             have my age precisely."
               How old is he now?
             Three Daughters and Two Sons
             An uncle visited his two nieces and three nephews.
               The first to greet him were little Johnny and his
274-275      Merry   Arithmetic

             sister Anne. The little chap proudly declared that he
             was twice as old as his sister. Then Nadine ran out to
             meet the uncle and her father said that both girls to-
             gether were twice as old as the boy.
                When Alexis came from school the father reckoned
             that both boys together are twice as old as both girls.
                The latest to come was Libby, she saw the guest and
             exclaimed happily: "Uncle, you just arrived on my
             birthday. Today I'm 21."
                "And you know what?" the father added, "It just
             occurred to me that my three daughters together are
             two times older than my sons."
                How old was each son and daughter?

             Two Trade       Unionists
             I remember hearing a conversation between two trade
               "So you've been a trade union member twice as long
             as me?"
               "Yes, exactly twice."
               "But last we met you said that you'd been a member
             three times longer."
               "Two years back? Then that was so, but now I've
             only been twice as long as you."
                How many years has each of them been a trade
             union member?

             How Many Games
             Three persons were playing draughts. They had played
             three games. How many games had each of them
Figure 243

             A snail was climbing up a 15-m tree. Each day it
             climbed 5 metres, but each night as it slept it slid back
             down 4 metres.
                How many days did it take the snail to reach the

             To the Town
             A farmer was travelling to a town. The first half of the
             route he went by train, 15 times faster than if he had
             gone by foot. However, the second half of the route he
             Merry     Arithmetic

             rode on an oxcart, at half the speed of a walker.
               How much time did he save as compared with
             walking all the way to the town?
Figure 244

             To the Village
             From a town to a village the road first goes uphill for
             8 kilometres, then 24 kilometres downhill. John went
             there on a bicycle and the non-stop journey there took
             him 2 hours 50 minutes. He also bicycled back, again
             non-stop, and spent 4 hours 30 minutes.
               How fast could John ride uphill and how fast

             Two     Schoolboys
             A schoolboy said to his mate, "Give me an apple, and
             ril have twice as many as you."
                "That would be unfair," replied the mate, "You give
             me one then we'll even."
                How many apples had each initially?

             Here is an insidious problem. A bound book costs
             2 roubles 50 kopecks. The book is 2 roubles more
             expensive than the binding.
               How much does the binding cost?

             The Cost of Buckle
             A belt with a buckle costs 68 kopecks. The belt costs 60
             kopecks more than the buckle.
               How much does the buckle cost?
             Merry   Arithmetic
Figure 245

             Casks of Honey
             In store there were seven casks brim full of honey,
             seven half-full ones, and seven empty casks, all
             belonging to three firms that wanted to divide both the
             honey and casks into equal shares.
                How can they divide without transferring the honey
             from one cask into another?
               If you think that various ways of doing so are
             available, indicate all those which occur to you.

             Postage    Stamps
             A man bought a 5 roubles worth of postage stamps of
             three kinds: 50-kopeck stamps, 10-kopeck stamps and
             1-kopeck stamps-100 pieces, all told.
                How many stamps of each kind did he buy?

             How Many        Coins?
             A customer got his change of 4 roubles 65 kopecks, all
             in 10 and 1-kopeck coins-in all, 42 coins.
               How many coins of each worth was he given?
               How many solutions has the problem?

             Socks and Gloves
             One box contains 10 pairs of brown and 10 pairs of
             black socks, another contains 10 pairs of brown and 10
             pairs of black gloves.
               How many single socks and gloves is it sufficient to
             take from each box to have any one pair of socks and
             Merry   Arithmetic

'9,          Book    Worm
Figure 246
             There are insects that feed on books eating their way
             through leaves and thus ruining the bulk of the book.
             One such book worm has eaten through from the first
             page of the first volume to the last page of the second
             volume on a bookshelf, as shown in the accompanying
                There are 800 pages in each volume so, how many
             pages has the worm ruined?
                The problem is not difficult but it has a catch.

             Spiders and Beetles
             A boy collects spiders and beetles in a box and he now
             has 8 insects in all. There are 54 legs in the box in
                How many spiders and how many beetles are there?

             Seven Friends
             A man has seven friends. The first visits him every
             night, the second every other night, the third every
             third night, the fourth every fourth night, etc., through
             to the seventh friend who comes every seventh night.

Figure 247

               Is it often the case that all the seven friends get to-
             gether there on the same night?

             The Same Problem Continued
             On those nights when the seven friends get together the
             host treats them to some wine, and all touch glasses in
               How many times do the glasses ring as they touch
             one another?
278-283                          Answers

                                Cats and Mats
This problem is solved in this way. Ask the question: How many more cats would be
needed to occupy all the places on the mats the second time than to get the situation
we had the first time? We can easily figure that out: in the first case one cat was left
without a place, whilst in the second case all the cats were seated and there were even
places for two more. Hence for all the mats to have been occupied in the second case
there should have been 1 + 2, i. e. three, more cats than there were in the first case.
But then each mat would have one more cat. Clearly there were three mats all in all.
Now we seat one cat on each mat and add one more to obtain the number of cats,
  Thus, the answer is four cats and three mats.

                                Sisters and Brothers
Seven: four brothers and three sisters. Each brother has three brothers and three
sisters; each sister has four brothers and two sisters.

                                How Many      Children?
Seven: six sons and one daughter. (The common answer is twelve, but each son would
then have six sisters, not one.)

The situation is very simple. Seating at the table were three, not four people: a grand-
father, his son, and his grandson, l l i e grandfather and his son are fathers, and the son
and grandson are sons.

                                Three Quarters of a Man
Note that the three-quarters-of-a-man is the last quarter of the team. So the whole of
the team is four times the three quarters of a man, i. e. three. In consequence, the team
consisted of three men.

                                How Old Are      They?
We know thus that the son is 7 times, and the grandfather 12 times, older than the
grandson. If the grandson were one year old, the son would be seven, and the grandfa-
ther 12. Adding the three together gives 20, which is exactly a fifth of the real figure. It
follows that in actuality the grandson is five, the son 35, and the grandfather 60.
  Check: 5 + 35 + 60 = 100.
                               A nswers

                               Who is Older?
Neither is: they are twins and each of them at the time is six.
  The calculation is simple: two years hence the boy will be four years older than he
was two years ago, and twice as old as he was then. Hence he was four years old two
years ago. Accordingly, now he is 4 + 2 = 6 years old.
  The girl's age is the same.

                               The Age of the Son
If now a son is one third the age of his father, then the father is older in years by
twice the son's age. Five years ago the father was, of course, also twice his son's
present age older than his son was then. On the other hand, since at that time the
father was four times older than the son, he was older by the triple then age of the
son. Consequently, the double present age of the son equals the triple then age of the
son. Thus, the son is now 11/2 times older than he was five years ago. It follows that
five years is a half of the son's previous age, and hence five years ago the son was 10,
and now he is 15 years old.
    Thus, the son is now 15 years and the father 45. Checking this, five years ago the
father was 40 and the son was 10, i.e. a quarter the father's present age.

                               His Age
Arithmetically the problem has a rather involved solution, but the situation simplifies
considerably if we draw on the services of algebra and set up an equation. We'll
denote the number of years we're after by x. The wit's age in three years will then be
x + 3, and three years ago, x — 3. We'll thus have
                              3 (x + 3) - 3 (x - 3) = x
The equation gives x = 18. So the witty person's age is at present 18 years.
   Let's check. In three years time he'll be 21 and three years ago he was 15. The
difference is
                               3 x 21 - 3 x 15 = 63 - 4 5 = 18,
which equals the present age of the man.

                               Three Daughters and Two Sons
We know that Johnny is twice as old as Anne, and Nadine and Anne together are
twice as old as Johnny. Accordingly, the sum of the ages of Nadine and Anne is four
times more than Anne's age. It follows directly that Nadine is three times as old as
   We also know that the ages of Alexis and Johnny combined are twice the combined
age of Nadine and Anne. But Johnny's age is double Anne's age, and the ages of
Nadine and Anne put together give the fourfold age of Anne. Accordingly, Alexis's age
plus the double age of Anne are equal to the eightfold age of Anne. Thus, Alexis is six
times older than Anne.
280-261                        Answers

  Lastly, as stated, the ages of Libby, Nadine, and Anne combined equal the sum of
the ages of Johnny and Alexis.
  For convenience, we'll compile the table:
                               Libby         21 years
                               Nadine        three times older than Anne
                               Johnny        two times older than Anne
                               Alexis        six times older than Anne
We can now say that 21 years plus the trebled Anne's age plus an Anne's age are
equal to the fourfold Anne's age plus the twelvefold Anne's age.
  Or, 21 years plus the fourfold age of Anne are equal to the sixteen-fold age of Anne.
  In consequence, 21 years are equal to the twelve-fold age of Anne and Anne is thus
21/12 = 1 3/4 years.
  •We can now easily determine that Johnny is 3 1/2, Nadine 5 1/4, and Alexis 10 1/2.

                               Two Trade    Unionists
One has been with the trade union for eight years, the other for four years. Two years
ago the first one had been with the trade union six years, the second two years, i.e.
three times as long. The problem is readily solvable using an equation.

                               How Many     Games?
The commonest answer is that each played once, ignoring the fact that three (and any
odd number in general) players cannot each play only once for who then did the third
player play? It takes two partners to have a game. If we denote the players by A, B,
and C, the three games will be
                               A with B
                               A with C
                               B with C
  We see that each played twice, not once:
                            A with B and C
                            B with A and C
                            C with A and B
So the answer is: each of the three played twice, although three games had been
played in all.
In 11 days. During the first 10 days the snail had crawled up 10 metres, 1 metre a day.
The next one day it climbed the remaining 5 metres, i. e. it reached the summit. (The
common answer is 15 days.)

                               To the Town
The farmer lost time, he did not save it. The second half of the route took as much
time as he would have spent travelling to the town on foot. He thus could not save
                               A nswers

time, he was bound to lose time.
   His loss amounted to 1/15 of the time required to cover half a route on foot.
                               To the Village
The solution of the problem follows from the following calculation:
   24 kilometres uphill and 8 kilometres downhill took 4 hours 30 minutes;
   8 kilometres uphill and 24 kilometres downhill took 2 hours 50 minutes.
   If we multiply out the second line by three, we obtain: 24 kilometres uphill and 72
kilometres downhill takes 8 hours 30 minutes.
   A bit of algebra gives that the bicyclist covers 64 kilometres downhill in 4 hours.
Hence, downhill he travelled at 64/4 = 16 kilometres an hour.
   We'll find in much the same way that he travelled uphill at 6 kilometres an hour.
Testing the answer is an easy exercise.
                               Two    Schoolboys
Transferring an apple balances out the number of apples, thus suggesting that one had
two apples more than the other. If we subtract one apple from the smaller number
and add it to the larger number, then the difference will increase by two and become
four. We know that then the larger number will be equal to double the smaller one.
Accordingly, the smaller number is 4 and the larger 8.
  Before the transfer one schoolboy had 8 — 1 = 7 apples, and the other 4 + 1 = 5
  Let's check whether or not the numbers become equal if we subtract an apple from
the larger and add it to the smaller:
                               7-1=6;       5 + 1 = 6.
  Thus, one schoolboy has 7 apples and the other 5 apples.

The off-the-cuff answer is usually: the binding costs 50 kopecks. But then the book
would cost 2 roubles, i.e. it would be only 1 rouble 50 kopecks more expensive than
the binding.
  The true answer is: the binding costs 25 kopecks, the book 2 roubles 25 kopecks
with the result that the book costs 2 roubles more than the binding.
                               The Cost of Buckle
Perhaps you've decided that the buckle costs 8 kopecks. If so, you're mistaken, as the
belt would then cost only 52 kopecks more than the buckle, not 60 kopecks more.
  The correct answer is the buckle costs 4 kopecks, then the belt costs 68 — 4 = 64
kopecks, i.e. 60 kopecks more than buckle.
                               Casks of Honey
The problem becomes a fairly easy exercise if we note that in the 21 casks bought
there were 7 + 3 1/2 = 10 1/2 caskfuls of honey. Each firm must then get 3 1 / 2 caskfuls
of honey and seven casks.
282-283                        A nswers

  We could divide them in the following two ways:
  First way
                                          f 3 full
                               1st firm i s 1 half-full
                                            1             f   T o t a l : 3 1/2 caskfuls of honey
                                           13 empty
                                        ( 2 full
                               2nd firmS 3 half-full          T o t a l : 3 1/2 caskfuls of honey
                                        12 empty
                                        f 2 full
                               3rd firm < 3 half-full             T o t a l : 3 1/2 caskfuls of honey
                                        I 2 empty

                                              3 full
  Second way                   1st   firm                 >
                                              1 half-full J   T o t a l : 3 1/2 caskfuls of honey
                                            . 3 empty
                                        [3 full
                               2nd firm ^ 1 half-full         T o t a l : 3 1/2 caskfuls of honey
                                          3 empty
                                        ( 1 full
                               3rd firmi 5 half-full }•       T o t a l : 3 1/2 caskfuls of honey
                                        11 empty

                               Postage       Stamps
There is only one answer: the customer bought
                                1 x 50-kopeck stamp
                               39 x 10-kopeck stamps
                               60 x 1-kopeck stamps
Really, there were 1 + 39 + 60 = 100 pieces all in all, and the total cost was 50 +
+ 390 + 60 = 500 kopecks.

                               How Many           Coins
The problem has four solutions:

                                                              I             II          III        IV

                               Roubles                         1             2           3          4
                               10-kopeck pieces               36            25          14          3
                               1-kopeck pieces                 5            15          25         35

                                                  Total       42            42          42         42

                               Socks and Gloves
Three socks will be enough, as two of them are bound to be of the same colour. But
with the gloves the situation is not that simple. These differ from one another not only
in their colour, but also in that half of them are right-handed and half left-handed.

Here 21 gloves will be sufficient. With a smaller number it might appear that all of
them would be right-handed, or left-handed for that matter (10 pairs of brown left and
10 pairs of black left).

                               Book       Worm
The common answer is that the worm went through 800 + 800 pages plus two covers.
But this is not so. Stand two books side by side as shown in Fig. 246, and see how
many pages there are between the first page of the first book and the last page of the
second book. You'll discover that there is nothing but two covers between them.
   Thus the book worm had only destroyed the two covers without touching their

                               Spiders and Beetles
To tackle the problem we should first of all remember from your nature lessons how
many legs beetles have and how many spiders have. In fact, the numbers are six and
eight, respectively. With this in view we suppose that the box only contains beetles,
either all told. Their legs would then add up to 6 x 8 = 48, six fewer than given in the
problem. Let's now try and replace one beetle with one spider. This will increase the
number of legs by two because the spider has two more legs.
   Clearly three such replacements will bring the total number of legs in the box up to
the desired 54. But in that case there will only be five beetles, the rest being spiders.
   The box thus contained five beetles and three spiders.
   Let's check: five beetles give 30 legs, the three spiders 24, the total being 30 + 24 =
 = 54, as required.
   The problem can also be solved in another way. We may start off assuming that the
box only contains spiders, eight of them. The number of legs will then be 8 x 8 = 64,
10 legs more than what was stated. Replacing one spider with one beetle reduces the
number of legs by two. We'll have to make five such substitutions in order that to
arrive at 54. Put another way, we'll retain only three spiders, with the rest being
replaced by beetles.

                               Seven Friends
You should easily twig that the seven can only come together in a number of days
that divides by 2, 3, 4, 5, 6, and 7. The smallest such number is 420.
  Consequently, the friends will only get together once every 420 days.

                                The Same Problem Continued
All of those present (the host and his seven friends) touch their glasses with those of
the remaining seven. The number of combinations 2 at a time totals 8 x 7 = 56. But
this counts each pair twice (e.g. the third guest with the fifth and the fifth with the
third are considered as different pairs). Hence the glasses ring 56/2 = 28 times.
284-285   Count

          Can You Count?
          The question might seem to be even insulting for
          a person more than three years old. Who can't? To
          utter the words "one", "two", "three" etc., in succession
          doesn't take much genius. And still I'm sure that you're
          not always equal to this seemingly simple task.
          Everything depends on what is to be counted. It's no
          problem to count, say, the nails in a box. But suppose
          the box contains screws as well as nails. It's required to
          find out how many of each there are. How could you
          go about it? Would you separate the heap into nails
          and screws and then count them?
             This sort of a problem comes up for a housewife
          when she has to count the washing for laundry. She
          first sorts the washing out: shirts go to one heap,
          towels to another, etc. And it's only after she had done
          this tedious job that she begins to count the items in
          each heap.
             That's what is called not knowing how to count!
          This way of handling dissimilar objects is utterly
          inconvenient, labour consuming and occasionally even
          completely impossible. It's all well and good if you
          have to count nails or washing: they are fairly easily
          sorted out into separate heaps. But try to place yourself
          into a forester's shoes who wants to count all the pines,
          spruces, birches and aspens in the same hectare. He
          cannot sort out all the trees according to their species.
          Well, should you first count all the pines, then all the
          spruses, then all the birches and then all the aspens?
          Would you go all round the whole area four times?
             Couldn't the job be done in a simpler way, perhaps
          by a single tour of the area? Yes, there is such a way
          and it has been used since time immemorial by
          foresters. Til illustrate its principle essence referring to
          our nails and screws.
             To count the nails and screw at one go, without
          sorting them out, get a pencil and a sheet of paper
          marked out as shown below:

                            Nails              Screws

                 Now begin counting. Take out of the box whatever
             comes first. If it's a nail you make a dash in "Nails", if
             it's a screw, mark a dash in "Screws". Take out
             a second piece and repeat the procedure, then a third,
             a fourth, etc., until the box is finished. In the end, in the
             "Nails" column you'll have as many dashes as there are
             nails in the box, and in the "Screws" column as many
             dashes as there are screws. It only remains to count up
             the dashes.
                 We could simplify the counting procedure. To do so
Figure 248   we should not just dispose our dashes one under
             another, but group them as shown in Fig. 248 with five
             dashes in each group.
                 It's convenient to arrange these squares into pairs,
             i.e. after the first 10 dashes begin a new row, then the
             third, and so on. The arrangement will be
             approximately as shown in Fig. 249.

Figure 249

                It's easy to count the dashes thus arranged: you      see
             at once that here we have three complete tens, one      five
             plus three dashes, i. e. 30 + 5 + 3 = 38.
                Other figures are also possible. So they often        use
             figures in which each complete square means               10
             (Fig. 250).

Figure 250

               When counting trees of different species in a forest
             area you should proceed in much the same way, only
             now you'll have four columns or lines, not two, lines
             being more convenient here. So you should begin with
             a sheet like this:
286-287        Count





Figure 251




                You'll end up with about what is shown in Fig. 251.
                It's a straightforward exercise here to work out the
                                  Pines 53     Birches 46
                                  Spruces 79   Aspens 37

   The same procedure is used by a medical worker
who counts under the microscope red and white blood
corpuscles in a blood specimen.
   Should you need to count the plants of various
species in a meadow you'll now know how to handle
the job and do it in the shortest time possible. First
write down the names of the plants found and allot
a line to each, leaving several lines for other plants you
may come across. Start off, for example, with an
arrangement like the one in Fig. 251, and proceed as if
it were the forest survey.

Why Count Trees in a Forest?
Why is it actually necessary to count the trees in
a forest? Some town dwellers even think that it's
impossible. In the novel Anna               Karenina    by
L.N. Tolstoy an agriculture expert, Levin, asked his
naive relative who wanted to sell his forest:
    "Have you counted the trees?"
    "How can one count trees?" he answers in
bewilderment. "Count sands or rays from distant
planets perhaps some lofty mind could..."
    "Oh yes? But the lofty mind of Ryabinin (a
merchant - the author) can. And no peasant will buy,
without counting."
    The trees in a forest are counted to assess the volume
of the wood in it. They count not all the trees in
a forest, but only those in a definite area, say a quarter
or half a hectare that is so chosen that the density,
composition, thickness and height of its trees were
representative of those in the entire forest. Selecting
a representative sample area requires, of course, a good
eye. The survey involves determining not only the
number of trees of each species, but also the number of
trunks of each gauge (say, 30 cm, 35 cm, etc.). The
report will therefore include more than four entries as
it's in our simplified example. Now you can imagine
how many times it would be necessary to go all round
the area if the trees were counted in some other way.
    You can see thus that counting is only a simple and
easy business when handling similar objects. When
handling dissimilar things the just described procedures
are needed.
288-289   Fast Reckoning

          (Simple tricks of mental        arithmetic)
          The following is a collection of simple and easily
          grasped tricks to speed up your mental arithmetic. If
          you want to master them you should realize that to be
          used fully they need to be approached conscientiously,
          not mechanically. But it pays to master them as they'll
          enable you to do calculations in your head without
          error, as with written calculations.

          Multiplication   by Simple      Number
          1. When multiplying by a simple number (for example,
          27 x 8) don't begin by multiplying the ones, as you
          would do in a written operation. First, multiply the
          tens of the multiplicand (20 x 8 = 160), then the ones
          ( 7 x 8 = 56) and add up the results (160 + 56 = 216).
             Further examples:
          34 x 7 = 30 x 7 + 4 x 7 = 210 + 28 = 238,
          47 x 6 = 4 0 x 6 + 7 x 6 = 240 + 42 = 282.
          2. It would also pay to remember the multiplication
          table up to 19 x 9 :

                           2    3    4     5   6     7   8    9
                     11    22   33   44   55   66   77   88   99
                     12    24   36   48   60   72   84   96 108
                     13    26   39   52   65   78 91     104 117
                     14    28   42   56   70   84 98     112 126
                     15    30   45   60   75   90 105    120 135
                     16    32   48   64   80   96 112  128 144
                     17    34   51   68   85   102 119 136 153
                     18    36   54   72   90   108 126 144 162
                     19    38   57   76   95   114 133 152 171

            Knowing the table you could multiply say 147 x 8 in
          your head as follows:
          147 x 8 = 140 x 8 + 7 x 8 = 1120 + 56 = 1176.
          3. If one of the numbers to be multiplied together is
          representable in the form of two factors, it may be
Fast   Reckoning

convenient to multiply in succession by three factors.
For example:
225 x 6 = 225 x 2 x 3 = 450 x 3 = 1350.

Multiplication     by Two-Digit        Number
4. This kind of multiplication can be made simpler by
reducing it to the conventional multiplication by
a simple number.
   When the multiplicand is simple, it and the multiplier
are interchanged and then the procedure of item 1 can
be followed. For example:
6 x 28 = 28 x 6 = 120 + 48 = 168.
5. When both multipliers are two-digit, one of them is
mentally broken down into tens and ones. For
29 x 12 = 29 x 10 + 29 x 2 = 290 + 58 = 348,
41 x 16 = 41 x 10 + 41 x 6 = 410 + 246 = 656,
(or 41 x 16 = 16 x 41 = 16 x 40 + 16 = 640 + 16 = 656).
It's more convenient to break the multiplier down into
tens and ones and so get smaller figures.
6. If the multiplicand or multiplier are more readily
representable in the head in the form of two simple
factors (e. g. 14 = 2 x 7), then this trick is used to reduce
one of the initial factors while increasing the other
accordingly (cf. item 3). For example:
45 x 14 = 90 x 7 = 630.

Multiplication     and Division by 4 and 8
7. To multiply in your head a number by 4, you double
the number twice. For example,
112 x 4 = 224 x 2 = 448,
335 x 4 = 670 x 2 = 1,340.
8. When multiplying by 8, the number is doubled three
times. For example,
217 x 8 = 434 x 4 = 868 x 2 = 1,736.
  Another way of multiplying mentally by 8 is to
multiply the multiplicand by ten and subtract double
the multiplicand (that is, multiply by 10 — 2 in the long
217 x 8 = 2 , 1 7 0 - 4 3 4 = 1,736.
290-291   Fast   Reckoning

          Or more convenient still:
          217 x 8 = 200 x 8 + 17 x 8 = 1,600 + 136 = 1,736.
          9. For a number to be mentally divided by 4, the
          number is halved twice. For example,
          76-1-4 = 38-1-2 = 19,
          236-1-4 = 118-1-2 = 59.
          10. To divide mentally by 8, the number is halved three
          times. For example,
          464 + 8 = 232-1-4 = 116+2 = 58,
          5 1 6 + 8 = 2 5 8 + 4 = 129+2 = 64 1/2.

          Multiplication       by 5 and 25
          11. Multiplying by 5 is actually multiplying by 10/2.
          Thus a zero is ascribed to the number and the result is
          divided by two. For example,
          74 x 5 = 7 4 0 + 2 = 370.
          243 x 5 = 2,430+2 = 1,215.

             If our number is even, it's more convenient to halve
          it first and then add the zero. For example,

          74 x 5 =           x 10 = 370.
          12. In the case of 25, a number is multiplied by 100/4,
          i.e. if the number is divisible by 4, it is divided by
          4 first and two zeros are then ascribed to the result.
          For example,
          72 x 25 = — x 100 = 1,800.
             But if the division yields a remainder, then if it's 1 we
          add 25 to the quotient, if 2 we add 50, and if 3 we add
          75. This follows from the fact that 100+4 = 25,
          2 0 0 + 4 = 50, and 3 0 0 + 4 = 75.

          Multiplication       by 11/2, 11/4, 2 1/2, 3/4
          13. When multiplying by 1 1/2, add to the multiplicand
          its half. For example,
          34 x 1 1/2 = 3 4 + 17 = 51.
          23 x 1 1/2 = 23 + 11 1/2 = 34 1/2 (or 34.5).
Fast   Reckoning

14. When multiplying by 1 1/4, add to the multiplicand
its quarter. For example,
48 x 1 1/4 = 48 + 12 = 60.
58 x 1 1/4 = 58 + 14 1/2 = 72 1/2 (or 72.5).
15. To multiply by 2 1/2, add to the doubled number
its half. For example,
18 x 2 1/2 = 36 + 9 = 45.
39 x 2 1/2 = 78 + 19 1/2 = 97 1/2 (or 97.5).
  Another technique consists in multiplying by 5 and
dividing by two:
18 x 2 1/2 = 9 0 ^ 2 = 45.
16. To multiply by 3/4 (that is, to find 3/4 of
a number), the number is multiplied by 11/2 and
divided by two. For example,
           3 0 + 15
30 x 3/4 =          = 22 1/2 (or 22.5).

  Another form of the technique consists in subtracting
from the multiplicand its quarter or adding to a half of
the multiplicand a half of the half of the multiplicand.

Multiplication     by 15, 125, 75
17. Multiplication by 15 is replaced by multiplying by
10 and then by 11/2 (because 10 x 1 1/2) = 15). For
18 x 15 = 18 x 1 1/2 x 10 = 270.
45 x 15 = 450 + 225 = 675.
18. Multiplication by 125 is replaced by multiplying by
100 and by 1 1/4 (because 100 x 1 1/4 = 125). For
26 x 125 = 26 x 100 x 1 1/4 = 2,600 + 650 = 3,250.
                                     4 700
47 x 125 = 47 x 100 x 1 1/4 = 470 + —      = 4700 +
 + 1,175 = 5,875.
19. Multiplication by 75 is replaced by multiplying by
100 and by 3/4 (because 100 x 3/4 = 75). For example,
                                        1800 -I- 900
18 x 75 = 18 x 100 x 3/4 = 1800 x 3/4 =      J       =
= 1,350.
292-291   Fast   Reckoning

          Note: Some of the above examples can be conveniently
                handled using the technique of item 6:
          18 x 15 = 90 x 3 = 270.
          26 x 125 = 130 x 25 = 3,250.

          Multiplication     by 9 and 11
          20. When multiplying by 9, add a zero to the number
          and subtract the multiplicand from the result. For
          62 x 9 = 620 - 62 = 600 - 42 = 558.
          73 x 9 = 730 - 73 = 700 - 43 = 657.
          21. When multiplying by 11, add a zero to the number
          and add the multiplicand to the result. For example,
          87 x 11 = 8 7 0 + 87 = 957.

          Division by 5, 1 1/2, 15
          22. To divide by 5 double the number and move the
          decimal point one place to the left. For example,

          6 8 + 5 = ^ - = 13.6.
          2 3 7 + 5 = — = 47.4.

          23. Dividing by 11/2 consists in doubling the number
          and dividing the result by 3. For example,
          36 + 1 1/2 = 7 2 + 3 = 24.
          53 + 1 1/2 = 106+3 = 35 1/3.
          24. Dividing by 15 consists in doubling the number
          and dividing the result by 30. For example,
          240+15 = 480+30 = 4 8 + 3 = 16.
          462+15 = 924+30 = 30 24/30 = 30 4/5 = 30.8.
          (or 924+30 = 308-^10 = 30.8).
          25. To square a number ending in 5 (e.g. 85) the
          number of tens (8) is multiplied by itself plus one (8 x
           x 9 = 72) and on the right of the result 25 is ascribed
          (in our example this yields 7,225). Some more examples:
          25 2 ; 2 x 3 = 6; 625.
          45 2 ; 4 x 5 = 20; 2,025.
Fast   Reckoning

145 2 ; 14 X 15 = 210; 21,025.
   The procedure follows from the formula:
(lOx + 5)2 = lOOx2 + lOOx + 25 = lOOx (x 4- 1) + 25.
26. This technique can also be applied to decimal
fractions ending in 5:
8.52 = 72.25; 14.52 = 210.25; 0.352 = 0.1225, etc.
27. As 0.5 = 1/2 and 0.25 = 1/4, the procedure of item
25 can also be used to square numbers ending in the
fraction 1/2:
(8 1/2)2 = 72 1/4.
(14 1/2)2 = 210 1/4, etc.
28. Mental squaring can often be simplified using the
(a ± b)2 = a2 + b2 ± lab.
For example,
41 2 = 40 2 + 1 + 2 x 40 = 1,601 + 80 = 1,681.
69 2 = 70 2 + 1 - 2 x 70 = 4,901 - 140 = 4,761.
362 = (35 + l) 2 = 1,225 + 1 + 2 x 35 = 1,296.
  The procedure is also convenient for numbers ending
in 1, 4, 6, and 9.

Calculations by Formula (a + b) (a — b) = a2 — b2
29. Let's multiply 52 x 48. We mentally represent the
multipliers as (50 + 2) (50 — 2) and use the formula:
(50 + 2) (50 - 2) = 502 - 2 2 = 2,496.
  This technique can be used whenever one multiplier
can be conveniently represented as a sum of two
numbers and the other as a difference of the same
numbers, e.g.
69 x 71 = (70 - 1) (70 + 1) = 4,899.
33 x 27 = (30 + 3) (30 — 3) = 891.
53 x 57 = (55 - 2) (55 + 2) = 3,021.
84 x 86 = (85 - 1) (85 + 1) = 7,224.
30. The procedure may conveniently be used             for
calculations of the following type:
I 1/2 x 6 1/2 = (7 + 1/2) (7 - 1/2) = 48 3/4.
II 3/4 x 12 1/4 = (12 - 1/4) (12 + 1/4) = 143 15/16.
294-291   Fast   Reckoning

          It Pays to Remember:
          37 x 3 = 111
          With this in mind we can easily carry         out   the
          multiplication of 37 by 6, 9, 12, etc.
          37 x 6 = 37 x 3 x 2 = 222.
          37 x 9 = 37 x 3 x 3 = 333.
          37 x 12 = 37 x 3 x 4 = 444.
          37 x 15 = 37 x 3 x 5 = 555, etc.

          7 x 11 x 13 = 1,001
          With this in mind we can easily carry         out   the
          multiplications of the following type:
          77 x 13 = 1,001.
          77 x 26 = 2,002.
          77 x 39 = 3,003, etc.
          91 x 11 = 1,001.
          91 x 22 = 2,002.
          91 x 33 = 3,003, etc.
          143 x 7 = 1,001.
          143 x 14 = 2,002.
          143 x 21 = 3,003, etc.
          We have only discussed the simplest and most
          convenient techniques of mental arithmetic. An
          inquiring mind can, through practice, work out further
          procedures to simplify calculations.
                    Magic Squares

                    The Smallest Magic       Square
                    Since time immemorial people have amused themselves
                    by constructing magic squares. The problem consists in
                    arranging successive numbers (beginning with 1) over
                    the cells of a divided square so that the numbers in all
                    the lines, columns and diagonals add up to the same
                      The smallest magic square has nine cells. It can easily
                    be shown by trials that a four-cell magic square is
                    impossible. The following is an example of a 9-cell
                    magic square:
Figure   252                                   4   3    8

                                               9   5    1

                                               2   7    6

                    In this square we might add up either 4 + 3 + 8, or 2 +
                     -I- 7 + 6, or 3 -(- 5 -I- 7, or 4 + 5 + 6, or any other line of
                    three numbers, the result is always 15. The result could
                    be envisaged beforehand, without constructing the
                    square as such: the three lines of the square should
                    contain all the nine numbers and they add up to
                    1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
                      On the other hand, this sum must clearly be equal to
                    thrice the sum of a single line. Hence for each line
                    45-^3 = 15.
                       Using the same argument we can determine in
                    advance the sum of the numbers in a line or column of
                    any magic square consisting of an arbitrary number of
                    cells. We only have to divide the sum of all its numbers
                    by the number of its lines.

                    Turns and Reflections
                    Haying constructed one magic square we can readily
                    derive its modifications, i.e. a series of new magic
                    squares. If, for instance, we have the square given in
                    Fig. 253, then by mentally turning it by 90° we'll
                    obtain another magic square (Fig. 254):
296-297            Magic   Squares

                   Figure 253            Figure   254

                     Further t u r n s - b y 180° and 270°-will give two more
                   modifications of the initial square.
                     Each of the new magic squares can in turn be
                   modified by reflecting it in a mirror. Figure 255 depicts
                   the initial square with one of its mirror reflections.

Figure 255

                     All the turns and reflections possible with the 9-cell
                   square yield the following versions (Fig. 256):

Figure 256 (1-3)

Figure 256 (4-8)

                     This is the complete collection of magic squares that
                   could be compiled of the first nine figures.

                   Bachefs      Method
                   Here's an ancient method of constructing odd magic
                   squares, i. e. squares with any odd number of cells: 3 x
             Magic   Squares

              x 3, 5 x 5, 7 x 7, etc. The method was suggested in the
             17th century by the French mathematician Claude-
             Gaspar Bachet (1581-1638). The method being suitable
             for the 9-cell square, it'll be convenient to begin
             discussing the method with this, the simplest case.
                So, having drawn a square divided into nine cells
             we'll write the numbers from 1 to 9 in succession
             arranging them in the oblique lines as shown in
             Fig. 257.

                                              2              6
Figure 257
                                              4              8
                                       : 7i
                                       . ;
                We transfer the numbers that appear to lie beyond
             the confines of the square into the square so that they
             join the lines at the opposite sides of the square. We
Figure 258
             thus obtain:

                                              9       5      1
                                              4       3      8

              Let's apply Bachet's technique to a 5 x 5                       square.
             We'll begin with the arrangement:
Figure 259                             : 5!
                                   i 4 • ! 10 i
                                 3       9     15
                                    2         8             14         20 |
                               r—(....                   —t—i
                               !1      7        13    19    ! 25 |
                               i f—•                        1 '
                                   6         12    18    24 '
                                       11             17          23
                                            i 16 '         i 22
                                            '     U   --
                                                  I 21
               It only remains now to bring the numbers outside
             the confines of the square into it. To do so, we'll need
             to bring the three-number arrangements from beyond
298-297                           Magic   Squares

                                  the square ("terraces") to the opposite sides of the
                                  square. This will give a 25-cell magic square (Fig. 260).
                                     The idea behind this simple technique is fairly
      3    16 9    22 15          complicated, though the reader can check it practically.
                                     Now that we have this magic square with 25 cells we
      20   8   21 14    2         can obtain its modifications by using turns and mirror
      7    25 13   1    19        reflections.
      24 12    5   18   6
                                  The Indian        Method
      11   4   17 10 23
                                  The Bachet (or "terrace") method is not the only
                                  approach to constructing squares with an odd number
                                  of cells. Worth mentioning among other techniques is
                                  a relatively easy procedure that is thought to have been
 30 39 48      1 10 19 28         devised in India about two thousand years ago. It can
                                  be briefly couched in six rules. Read them carefully and
 38 47     7   9   18 27 29       then see them applied to a magic square with 49 cells
 46   6    8   17 26 35 37        (Fig. 261).
                                     1. In the middle of the upper line we write 1, and at
  5   14 16 25 34 36 45           the bottom in the next column on the right we write 2.
 13 15 24 33 42 44           4       2. We write the next numbers successively along the
                                  diagonal upwards and to the right.
 21 23 32 41 43         3 12         3. Having reached the right edge of the square we go
 22 31 40 49       2    11 20     over to the extreme left cell in the next line up.
                                     4. Having reached the upper edge of the square we
                                  go over to the lowest cell of the next column to the
                                     Note: Having reached the upper right corner cell we
                                             go over to the leftmost lower corner cell.
                                     5. Having reached an occupied cell we skip over it.
                                     6. If the last occupied cell belongs to the lowermost
                                  line, we proceed from the uppermost cell in the column.
                                     Observing these rules we can quickly construct magic
                                  squares with any odd number of cells.
                                     If the number of cells in the square doesnt divide into
                                  3 we may begin the square following a rule other than
 32 41 43      3   12 21 23          We may start from any cell along the diagonal line
 40 49     2   11 20 22 31        passing between the middle cell of the leftmost column
                                  and the middle cell of the uppermost line. All the other
 48    1   10 19 28 30 39         numbers are placed according to rules 2-5.
  7    9   18 27 29 38 47            It is possible therefore to construct several squares by
                                  this method. By way of example, we provide the
  8   17 26 35 37 46         6    following 49-cell magic square (Fig. 262).
 16   25 34 36 45       5    14      Exercise. Use the Indian method to construct several
                                  magic squares with 25 and 49 cells. Obtain other
 24 33 42 44        4   13   15   squares by turns and mirror reflections.
                               Magic   Squares

    <•»                        Squares with an Even Number of Cells
Figure   263
                               These magic squares can't be constructed using any
                               common or convenient rule. There is one relatively
                               simple procedure for even squares, the number of
                               whose cells is divisible by 16. This means that one side
         0                     of these squares has a number of cells that is a multiple
                               of 4, i.e. it consists of 4, 8, or 12, etc., cells.
                        0         We'll now agree as to what we'll call "opposite" cells.
               X               As an example, Fig. 263 presents two pairs of opposite
                               cells: one marked by crosses, the other by circles.

                              We see that if a cell lies in the second line from the
                           top and fourth from the left, then the respective
Figure 264                 opposite cell will lie in the penultimate line and fourth
                           from the right. (The reader is recommended to practise
    1 2 3 4 5 6 7 8 finding some other "opposite" cells.) Note that the
                           opposites of cells in a diagonal also lie on the same
    9 10 11 12 13 14 15 16 diagonal.
   17 18 19 20 21 22 23 24    We'll explain the procedure referring to 8 x 8 square.
                           To begin with, we'll place all the numbers from 1 to 64
   25 26 27 28 29 30 31 32 in the cells in succession (Fig. 264).
   33 34 35 36 37 38 39 40    All the diagonal lines in the resultant square have the
                           same sum -260, just what we want for a 8 x 8 magic
   41 42 43 44 45 46 47 48 square (Check!). But the lines and columns of the
                           square give different sums. So, the upper line adds up
   49 50 51 52 53 54 55 56
                           to 36, i.e. 224 less than required, and the last line adds
   57 58 59 60 61 62 63 64 up to 484, i.e. 224 more than required. Noting that
                           each number in the last line is 56 more than the
                           number in the same column but in the first line and
                           that 224 = 4 x 56, we come to the conclusion that the
                           sums of these two lines can be equalized if we replace
                           a half of the numbers in the first line by the
                           corresponding number in the last line. For instance, the
                           numbers 1, 2, 3, and 4 are replaced by 57, 58, 59, and
                           60, respectively.
                              What has been said about the first and eighth lines is
                           also true of the second and seventh, the third and sixth,
                           and in general for any pair of lines equidistant from
                           their respective extremum lines (i. e. first and last). After
                           the numbers in all the lines have been interchanged
                           we'll obtain a square in which lines have equal sums.
                              It is, however, necessary that the columns, too, give
                           the same sum. With the initial arrangement we could
                           have achieved this by the same kind of exchange that
                           we used with the lines. But after rearranging the lines,
                           the situation has become more complex. To identify the
                           numbers to be exchanged we make use of the following
300-297                              Magic    Squares

                                     technique, which could have been applied from the very
                                     beginning. Then, instead of a double rearrangement (of
                                     the lines and columns) we exchange "opposite"
Figure 265                           numbers. But this rule is not sufficient in itself,
                                     however, since we've found that only a half of the
 1X 2        3   4X 5x 6    7    8 X numbers need to be exchanged, the remaining numbers
                                     staying in their previous places. Which of the opposite
 9 X 10x 11 12 13 14 15 X        16X pairs then are to be exchanged?
 17 18* 19 x 20 21 22 x 23 x     24     The following four rules are an answer to this
 25 26 27* 28 X 29X 30 x 31      32     1. We divide the magic square into four squares as
 33 34 35 36         37 38 39     40 shown in Fig. 265.
 41 42 43 44 45 46 47 48     2. In the upper left corner we mark with crosses
 49 50 51 52 53 54 55 56 a half of all the cells so that each column and line has
                          exactly one half of its cells marked. This can be done in
  57 58 59 60 61 62 63 64 a variety of ways, an example is shown in the above
Figure 266                   3. In the right upper square we mark with crosses
                          the cells that are symmetrical about those marked in the
  64 2 3 61 60 6 7 57 upper left corner.
                             4. We now only have to replace the numbers in the
  56 55 11 12 13 14 50 49 cells marked by those in the opposite cells.
                             As a result of the permutation we will have obtained
  17 47 46 20 21 43 42 24
                          a 64-cell magic square presented in Fig. 266.
  25 26 38 37 36 35 31 32    We could, however, use many other ways of marking
                          the cells in the left upper square so that rule 2 would
  33 34 30 29 28 27 39 40
                          be fulfilled, for example, as shown in Fig. 267.
  41 23 22 44 45 19 18 48
                                     Figure   267
 16 15 51 52 53 54 10            9
 8   58 59       5   4   62 63   1

                                       The reader will undoubtedly find many other ways of
                                     arranging the crosses in the cells of the upper left
                                        Using then rules 3 and 4 we can readily derive other
                                     magic squares with 64 cells.
Magic   Squares

  Arguing along these lines we can construct magic
squares consisting of 12 x 12, 16 x 16, etc., cells.
  The reader can do this on his own.

Whence the Name?
The first recorded evidence of the magic square comes
from an ancient oriental book referring to 4,000-5,000
B. C. Indians in ancient times had a better
understanding of magic squares, from where the
passion for magic squares was taken over by ancient
Arabs, who would assign mysterious qualities to these
combinations of numbers.
   In medieval Western Europe magic squares were the
stock-in-trade of representatives of pseudosciences, such
as alchemistry and astrology. It is from the medieval
superstitious perceptions that these squares have
derived their unusual name-"magic". Astrologists and
alchemists believed that a magic square drawn on
a piece of wood was able to deliver a man from

                         *   *   *

   The construction of magic squares is not just
a pastime. Many famous mathematicians have been
interested in their theory and it has been applied in
some of the important problems of mathematics. So,
there is a way of solving sets of equations in many
unknowns that uses results from the theory of magic
302-303           Arithmetic Games and Tricks

                  A Chain of 28 Bones. Why is it possible to have
                  a continuous chain of 28 bones (a domino piece is also
                  sometimes called a bone) constructed without breaking
                  the rules of the game?

                  Beginning and End of the Chain. The 28-bone chain
                  ends in 5 points. How many points are there at the
                  other end?

                  Trick with Dominoes. Your friend takes one of the
                  dominoes and tells you to build a continuous chain out
                  of the remaining 27 bones. He insists that it's always
                  possible whichever bone he takes. He leaves you on
                  your own and goes to another room.
                    You begin working and see that your friend is right:
                  the 27 bones produced a chain. What is more
                  surprising is that your friend, although remaining in the
                  other room, calls out the number of points at each end
                  of the chain.
                    How does he know? Why is he confident that any 27
                  bones can produce a continuous chain?

                  Frame. Figure 268 shows a square frame made from
                  dominoes whilst observing the rules. The sides of the
                  frame may be equal in length but not in the total
                  number of points: the upper and left sides contain 44
                  points each and the other two sides contain 59 and 32.
Figure 268
             Arithmetic   Games and TYicks

               Can you produce a square frame such that each side
             contains the same sum total of points-44?

             Seven Squares. We can select four bones so that these
Figure 269
             will form a square with the same sum of points on each
             side. An example is given in Fig. 269 in which the
             points on each side add up to 11.
                Using a complete set of dominoes, can you build
             seven such squares? They do not have to have the same
             common sum of points of their sides.

             Domino Magic Squares. Figure 270 shows a square of
Figure 270   18 dominoes that is remarkable in that the sum of the
             points on any of its lines (be it longitudinal, transverse
             or diagonal) is the same, namely 13.
                You are asked to construct several such 18-bone
             magic squares, but now with another line sum. For an
             18-bone square 13 is the smallest sum whilst 23 is the

             Domino Progression. In Fig. 271 you see six bones
             arranged according to the rules of the game but note
             that the number of points on each (both halves)
             increases by one. Beginning with 4, the series consists of
Figure 271   the following numbers of points 4, 5, 6, 7, 8, and 9.
               A series of numbers increasing (or decreasing) by the
             same amount each time is called an "arithmetic
             progression". In our case each number is one more
             than the previous one, but the difference may be

               Try to construct some other 6-bone progressions.

             The Fifteen Puzzle*
             The well-known tray with 15 numbered square counters
             has a curious history few people even suspect of. We'll
             recall it in the words of W. Arens, a German
             mathematician and investigator:
               "About half a century ago, in the late 1870s, the
             Fifteen Puzzle bobbed up in the United States; it
             spread quickly and owing to the uncountable number
             of devoted players it had conquered, it became
             a plague.
               "The same was observed on this side of the Ocean, in
                * Other    names   are the   Boss   Puzzle,   Jeu   de   Taquin,   and
304-305   Arithmetic   Games and   Thicks

           Europe. Here you could even see the passengers in
           horse trams with the game in their hands. In offices and
           shops bosses were horrified by their employees being
           completely absorbed by the game and they were forced
           to ban the game during office and class hours. Owners
           of entertainment establishments were quick to latch
          onto the rage and organized large contests. The game
           had even made its way into solemn halls of the German
           Reichstag. 'I can still visualize quite clearly the grey-
          haired people in the Reichstag intent on a square small
          box in their hands,' recalls the geographer and
          mathematician Sigmund Giinter who was a deputy
          during the puzzle epidemic.
             "In Paris the puzzle flourished in the open air, in the
          boulevards, and proliferated speedily from the capital
          all over the provinces. A French author of the day
          wrote, 'There was hardly one country cottage where
          this spider hand't made its nest lying in wait for
          a victim to flounder in its web.'
             "In 1880 the puzzle fever seems to have reached its
          climax. But soon the tyrant was overthrown and
          defeated by the weapon of mathematics. The
          mathematical theory of the puzzle showed that of the
          many problems that might be offered only a half were
          solvable, the other half were impossible, however
          ingenious the technique applied to solve them.
             "It thus became clear why some problems wouldn't
          yield under any conditions and why the organizers of
          the contests had dared offer such enormous rewards for
          solving the problems. The inventor of the puzzle took
          the cake in this respect suggesting to the editor of
          a New York newspaper that he publish an unsolvable
          problem in the Sunday edition with a reward of 1,000
          dollars for its solution. The editor was a little reluctant
          so the inventor expressed his willingness to pay his own
          money. The inventor was Sam Loyd. He came to be
          widely known as an author of amusing problems and
          a multitude of puzzles. Curiously enough, he failed to
          patent his Fifteen Puzzle in the USA. According to the
          regulations, he had to submit a "working model" so
          that a prototype batch could be manufactured from it.
          He posed the problem to a Patent Office official, but
          when the latter enquired if it were solvable, the answer
          was 'No, it is mathematically impossible'. The official
          therefore reasoned: 'In which case there can't be
          a working model and without a working model there
          can be no patent.' Loyd was satisfied with the decision.
                 304-305   Arithmetic   Games and Thicks

                 He would perhaps have been more insistent had he
                 foreseen the unprecedented success of his invention*."
                   We'll continue the story of the puzzle using the
                 inventor's own words:
                 Figure   272           1    2    3        4

                 The normal             5    6    7        8
                 arrangement of
                 counters (I)           9    10   11   12

                                        13   14   15

                    "The old dwellers of the realm of aptitude will
                 remember how in the early 1870s I made the whole
                 world rack its brains over a tray of movable counters,
Figure 273       that came to be known as the Fifteen Puzzle. The
                 fifteen counters were arranged in order in the tray with
                 only 14 and 15 counters inverted as shown in the
The unsolvable   accompanying illustration (Fig. 273). The puzzle was to
case (II)        get the counters into the normal arrangement by
                 individually sliding them so that the 14 and 15 were
                     "The 1000-dollar reward offered for the first correct
                 solution remained unretrieved although everybody was
                 busy on it. Funny stories were told of shop-keepers
                 who forget for this reason to open their shops, of
                 respectful officials who stood throughout the night
                 under a street lamp seeking a way to solve it. Nobody
                 wanted to give up as everyone was confident of
                 imminent success. It was said that navigators allowed
                 their ships to run aground, engine drivers took their
                 trains past stations, and farmers neglected their
                                                  *    *       *

                   We'll now introduce the reader to the beginnings of
                 the game. In its complete form it's very complicated
                 and closely related to one of the branches of higher
                 algebra (the theory of determinants). We'll just confine
                 our discussion to some of the elements as presented by
                 W. Arens.
                   The task of the game is normally as follows: using
                 successive movements made possible by the presence of
                 the blank space the arbitrarily arranged squares should
                 be brought to the normal arrangement, i. e. the counters
                 are in numerical order with the 1 in the upper left
                    * The episode was used by M a r k              Twain in his novel   The
                 American Claimant.

306-307   Arithmetic   Games and   Tricks

          corner, followed by the 2 on the right, then the 3 and
          the 4 in the upper right corner; in the next row there
          should be from left to right the 5, 6, 7, 8, etc., with the
          blank space ending up back in the lower right corner.
          The normal arrangement is given in Fig. 272.
             Now think of an arrangement with the 15 counters
          scattered arbitrarily. A number of movements can
          always bring the 1 to the place occupied by it in the
          figure. In exactly the same way we can, without touch-
          ing counter 1 move counter 2 to the adjacent place on
          the right. Next, without touching either the 1 or 2, we
          can move the 3 and 4 to their normal places. If these
          occasionally are not in the two last columns, we can
          bring them there and through a number of movements
          achieve the arrangement sought. Now the upper line is
          in the normal order and we'll leave it as it is in later
          manipulations. In the same way we'll also bring the
          second line into the normal order. We'll easily find that
          it's always possible. Further, within the space of the
          two last lines we'll need to arrange counters 9 and 13,
          which is always possible, too. It now only remains to
          arrange a small patch of six spaces, of which one is free
          and the other five are occupied by the 10, 11, 12, 14,
          and 15, arbitrarily arranged. Within this patch we can
          always bring the 10, 11, and 12 into the normal
          arrangement. This done, the 14 and 15 will be arranged
          in the last line either in the normal or inverted order
          (Fig. 273). This procedure, which the reader can easily
          test in practice, will always yield the following result.
              Any initial arrangement can be brought into either
          the Fig. 272 (/) form or the Fig. 273 (II) form:
              If an arrangement, we'll denote it by S, can be
          brought to /, then the opposite is clearly possible, i. e.
          I can be brought to S. After all, the movements are all
          reversible. If, for instance we can push the 12 in
          arrangement into the blank space, then we can always
          restore the previous arrangement by the reverse move.
              We thus have two series of arrangements such that
          the arrangements in one series can be brought to
          normal arrangement /, and those in the other series can
          be brought to arrangement II. Conversely, from the
          normal arrangement we can obtain any arrangement in
          the first series, and from II any arrangement in the
          second series. Lastly, any two arrangements in the same
          series are transferable one to the other.
              Could we go further and combine the two

                                   Arithmetic   Games and   Tricks

                                   arrangements? We could rigorously prove (we are not
                                   going to here) that these arrangements cannot be
                                   interchanged, however many moves are used. Therefore,
                                   the formidable variety of arrangements break down
                                   into two separate series: (1) those that can be brought
                                   into the normal arrangement, and (2) those that can't
                                   and it was for these arrangements that the enormous
                                   rewards were promised.
                                      How are we to know whether or not a given
                                   arrangement belongs to the first series? An example
                                   will clarify this.
Figure 21A                            Let's consider the arrangement shown in Fig. 274.
                                   The first line is in perfect order, as is the second save
                                   for the last counter (9). Counter 9 comes before 8. This
               1         3    4
                                   sort of violation of the order is called inversion.
               5    6    7    9    Concerning counter 9 we'll say that here we have one
                                   inversion. Further examination reveals that the 14
               8    10        12
                                   precedes three counters (12, 13, and 11), thus giving
               13   11   15        three inversions (14 before 12, 14 before 13, and 14
                                   before 11). This amounts to 1 + 3 = 4 inversions. Fur-
                                   ther, the 12 precedes the 11 and the 13 precedes the 11.
                                   This adds two more inversions bringing the total to six.
                                   This procedure is used to determine the total number
                                   of inversions for any arrangement with the blank space
                                   in the lower right corner. If, as in the case in hand, the
                                   total number of inversions is even, then the
                                   arrangement can be brought to the normal one; in
                                   other words, it's solvable. If the number of inversions is
                                   odd, the arrangement belongs to the second series, i. e. it
                                   is an insolvable arrangement.
                                      Thanks to the new light shed on the puzzle by
                                   mathematics the earlier morbid passion that was shown
                                   for the game is now unthinkable. Mathematics has
                                   produced an exhaustive explanation of the game, one
                                   that leaves no loophole. The outcome of the game is
                                   dependent not on chance nor on aptitude, as in other
                                   games, but on purely mathematical factors that
                                   predetermine it unconditionally.
                                      We'll now consider some of the solvable problems
                                   with the game that were produced by the resourceful
Figure   275
                    1    2    3
                                   Problem I. Starting off the arrangement in Fig. 273
               4    5    6    7    bring the counters into the numerical order with the
               8    9    10   11
                                   blank space in the upper left corner (Fig. 275).
               12   13   14   15   Problem II. Starting off with the arrangement           in
308-309      Arithmetic   Games and   Tricks

             Fig. 273 turn the tray 90° to the right and obtain the
Figure 276   arrangement of Fig. 276.

             Problem III. By moving the counters according to the
             rules turn the tray into a magic square, i. e. arrange the
             counters so that the sum of the numbers in all
             directions is 30.

             The "11" Game
             This is a game for two. Eleven matches (or other
             objects) are placed on a table. One player takes one,
             two or three matches, just as he likes. Then the other
             also takes one, two or three matches. Now again the
             first, and so on. It's forbidden to take more than three
             matches at a time. He who takes the last match loses.
                How must you play so that you can always win?

             The "15" Game
             This game is not to be confused with the Fifteen
             Puzzle. It's more like the well-known "noughts and
             crosses" game. It's played by two people taking turns.
             Each player writes a number from 1 to 9 in one of the
             cells of the network shown below.

                Each player selects his cell so that his opponent
             couldn't complete a row of three figures (the row may
             be transverse or diagonal) that add up to 15.
                The player completing such a row or filling in the
             last cell of the network is the winner.
                Is there any way of winning the game with certainty?

             The "32" Game
             First 32 matches are arranged on a table. Two people
             play alternately. The beginner draws one, two, three, or
             four matches, and then the other player also takes as
             many matches as he chooses, but again not more than
             four. And so on. The player taking the last match wins.
               The game, you see, is very simple but it is curious in
             that the beginner can always win if he plays correctly.
                Could you indicate the "right" way to win?
             304-305   Arithmetic   Games and   Thicks

             The Reverse of the Last Game
             The previous game can be modified so that the player
             taking the last match loses, not wins.
               How must you play then to win with certainty?

             The "27" Game
             The game is similar to the previous ones. It's also
             played by two people and also requires that the players
             alternately take no more than four matches. But the
             object of the game is different: the winner is the one
             who ends up with an even number of matches.
                The beginner here is at advantage, too. He can so
             calculate his draws that he always will win.
                What is the secret of his fail-safe strategy?

             The Reverse of the Last Game
             The object of the "27" game can be reversed so that the
             winner is the one ending up with an odd number of
               In this case, what is the fail-safe procedure?

             Arithmetic    TYavel
             Several people may take part in this game. You'll need:
               (1) a board (of cardboard),
               (2) a die (of wood),
               (3) several counters (as many as there are players).

Figure 277
310-311                      Arithmetic   Games ana   Tricks

                                The board is a cardboard square, preferably a large
                             one, divided into 10 x 10 cells that are numbered from
                             1 to 100 as shown in Fig. 277.
                                The die, about 1 cm on side, is made of wood. The
                             faces are sandpapered and numbered from 1 to 6 (or
                             marked with points as dominoes).
                                The counters may be various coloured disks, squares,
                                Taking turns, the players throw the die. If the die
                             shows, say, 6, the player moves his counter 6 squares
                             forward, his next throw takes his counter forward by as
                             many cells as there are points on the die. When the
                             player's counter comes to a cell where an arrow begins,
                             the counter must follow the arrow to its end either
                             forwards, or backwards.
                                The player whose counter first reaches 100 is the

                             Think of a Number
                             Think of a number, follow the procedure given below,
                             and Fll guess the result of your calculations.
                               Should the result differ, check through your
                             calculations since you will have been in error, not I.

No. 1                                         Add 14;
The number must be less                       Subtract 8;
than 10 though not zero                       Cross out the first digit;
Multiply it by 3;                             Divide by 3;
Add 2;                                        Add 10.
Multiply by 3;                                           The result is 12
Add the number thought
of;                                           No. 3
Cross out the first digit;
Add 2;                                        The number must be less
Divide by 4;                                  than 10 though not zero
Add 19.                                       Add 29 to it;
                                              Discard the last digit;
          The result is 21                    Multiply by 10;
                                              Add 4;
                                              Multiply by 3;
No. 2                                         Subtract 2.
The number must be less                                        The result is 100
than 10 though not zero
Multiply it by 5;
Multiply by 2;
                             316-313    Arithmetic   Games and   Tricks

No. 4                               Subtract it from 130;
The number must be less than        Add 5;
10 though not zero                  Add the number thought of;
Multiply it by 5;                   Subtract 120;
Multiply by 2;                      Multiply by 7;
Subtract the number thought of;     Subtract 1;
Add up the digits;                  Divide by 2;
Add 2;                              Add 30.
Square it;
Subtract 10;                                         The result is 40
Divide by 3.
                 The result is 37
                                    No. 8
                                    Any number (besides zero)
No. 5                               Multiply it by 2;
The number must be less than        Add 1;
10 though not zero                  Multiply by 5;
Multiply it by 25;                  Discard all the digits but the last;
Add 3;                              Multiply it by itself;
Multiply by 4;                      Add up the digits.
Cross out the first digit;
Square it;                                                  The result is 7
Add up the digits;
Add 7.
                The result is 16    No. 9
                                    The number must be less than 100
                                    Add to it 20;
                                    Subtract from 170;
No. 6                               Subtract 6;
The number must have two digits     Add the number thought of;
Add 7;                              Add up the digits;
Subtract it from 110;               Multiply it by itself;
Add 15;                             Subtract 1;
Add the number thought of;          Divide by 2;
Divide by 2;                        Add 8.
Subtract 9;
Multiply by 3.                                            The result is 48
               The result is 150

                                    No. 10
                                    The number must be three digits long
No. 7                               Write the same number on its right;
The number must be less than 100    Divide by 7;
Add 12 to it;                       Divide by the number thought of;
312-313                     Arithmetic   Games and   Tricks

Divide by 11;                                 Multiply by 2;
Multiply it by 2;                             Multiply by 2;
Add up the digits.                            Add the number thought of;
                                              Add the number thought of;
   The result is 8                            Add 8;
                                              Discard all the digits but the last;
                                              Subtract 3;
No. 11                                        Add 7.
The number must be less than 10
Multiply it by 2;                                                      The result is 12

                            Guessing a Three-Digit            Number
                           Think of a three-digit number. Leave aside the last two
                           digits and double the first one. Add 5 to the result, then
                           multiply by 5, add the second digit and multiply by 10.
                           Add the third digit to the new result and tell me what
                           you've arrived at. Fll immediately guess the number
                           you've thought of.
                              Let's take an example. Say your number is 387.
                              It goes through the following sequence of operations.
                              You double the first digit: 3 x 2 = 6;
                              Add 5: 6 + 5 = 11;
                              Multiply by 5: 11 x 5 = 55;
                              Add the second digit: 55 + 8 = 63;
                              Multiply by 10; 63 x 10 = 630;
                              Add the third digit: 630 + 7 = 637.
                              So you tell me the final result (637) and I tell you the
                           initial number. Explain how.

                            Another Number Trick
                           Think of a number;
                             Add 1;
                             Multiply by 3;
                             Add 1 again;
                             Add the number thought of;
                             Tell me the result.
                             When you tell me the result I subtract 4 from it,
                           divide the difference by 4 and obtain the number you
                           thought of.
                             For instance, suppose you thought of 12.
                             Add 1, we get 13.
                             Multiplied by 3, we get 39.
                             Added 1, we get 40.
                             Added the number thought of: 4 0 + 12 = 52.
                             When you tell me the number, 52, I subtract 4 from
318-313            Arithmetic   Games and   Tricks

it, and divide the difference, 48, by 4. I thus get 12, the
number you thought of.
    How does the procedure work?

Guessing the Crossed-Out Digit
Ask your friend to think a multidigit number and then
ask him to do the following:
  Write the number down;
  Transpose its digits in an arbitrary order;
  Subtract the smaller number from the larger;
  Cross out one of the digits (but not a zero);
  Name the remaining digits in any order;
  You will theti tell your friend the crossed-out digit.
Example. Your friend thought of 3857.
He performed the following:
8735 - 3857 = 4878.
   Your friend crosses out the 7 and tells you the
remaining digits in the following order, say:
8, 4, 8.
From these digits you can determine the crossed digit.
  How can this be done?

Guessing the Day and Month of Birth
Get your friend to write down the day and month of
his (or her) birth and to carry out the following
   Double the day;
   Multiply by 10;
   Add 73;
   Multiply by 5;
   Add the serial number of the month of birth.
  When he (or she) tells you the final result of his (or
her) calculations, you can tell him (or her) the day and
month of his (or her) birth.
Example. Suppose your friend was born on the 17 of
August, i.e. on the 17th of the 8th month. He does the
  17 x 2 = 34;
  34 x 10 = 340;
  340+73 = 413;
314-315   319-313   Arithmetic   Games and   Tricks

           413 x 5 = 2065;
          2065 + 8 = 2073.
            Your friend tells you the number 2073 and you tell
          him his birthday.
            How can you do this?

          Guessing Someone's        Age
          You can guess the age of a friend if you ask him (or
          her) to do the following:
                 Write down side by side any two digits that differ
                 in more than 1;
                 Write any digit between them;
                 Reverse the order of the three-digit number
                 Subtract the smaller number from the larger;
                 Reverse the digits of the difference;
                 Add the result to the difference;
                 Finally add his age to the sum.
                 Your friend tells you the final result of the
                 and then you can tell him his age.
          Example. Your friend is 23. He performs the following:
           5 7 2 - 2 7 5 = 297;
           297 + 7 9 2 = 1089;
          1089 + 23 = 1112.
            The number 1112 is the final result and from it you
          determine the age. How?

          How Many Sisters? How Many                  Brothers?
          You can guess how many brothers and sisters your
          friend has, if you ask him to do the following:
             Add 3 to the number of brothers;
             Multiply by 5;
             Add 20;
             Multiply by 2;
             Add the number of sisters;
             Add 5.
             The friend tells you the final result of his
          computations and you can tell him how many brothers
          and sisters he has.
320-313                Arithmetic   Games and   Tricks

Example. Your friend has four brothers and seven
sisters. He thus does the following:
   4 + 3 = 7;
  7 x 5 = 35;
 35 + 20 = 55;
 5 5 x 2 = 110;
110 + 7 = 117;
117 + 5 = 122.
  The friend tells you the number 122 and you can tell
him how many brothers and sisters he has.
  How can you do this?

Trick with a Telephone Directory
Here is another impressive trick. Get your friend to
write down any number with three different digits.
Suppose he writes 648. Ask him to reverse the digits in
the number he has chosen and subtract the smaller one
from the larger*. He will thus write:
~ 648

Ask to rearrange the digit of the difference in the
reverse order and add both numbers up. Your friend
will write:

   These calculations should be done in secret so your
friend thinks that the final result must be unknown to
   Now give your friend a telephone directory and ask
him to open it on the page whose number is equal to
the first three digits of the final result. He does so and
waits for further instructions. You then ask him to
count the telephone subscribers (down from the top or
up from the bottom) until he gets to the one given by
   * If the difference is a two-digit number (99), it is written with
a zero in front (099).
316-313   Arithmetic    Games and     Tricks

          the last digit of the number (1089). He thus finds the
          ninth subscriber and you tell him the name of the man
          and his telephone number.
            This naturally amazes your friend: he selected
          a number at random and you can tell him the
          subscriber's name and number.
            What is the trickery here?

          Guessing Domino Points
          The trick is arithmetic, based on calculation.
            Let your friend put a domino piece into his pocket.
          You promise to guess the number of points if he makes
          some simple calculations. Let his bone be the 6-3. Ask
          him to double one of the numbers (e.g. 6)
          6 x 2 =      12,

          and add 7
          12 + 7 = 19.
          Ask him to multiply the result by 5
          19 x 5 = 95
          and to add the other number of points of the domino
          piece (i.e. 3)
          95 + 3 = 98.
            He tells you the final result and you in your head
          subtract 35 to find the points on the piece: 98 — 35 =
          = 63, i. e. the piece was the 6-3.
            Why is it so and why one must always subtract 35?

          Formidable         Memory
          Conjurers sometimes amaze the public by their striking
          memory: they memorize long series of words, numbers,
          etc. Each of you can also surprise friends with such
          a trick.
             On 50 small paper cards write the numbers and
          letters shown below: a long number and in the left
          corner a letter or a combination of a letter and a figure.
          Distribute these cards among your friends and claim
          that you remember exactly which number is on which
          card. They need only tell you the number of the card
          and you'll immediately tell them the number written on
          it. You are told, say, "E4" and you can say at once
322-313                  Arithmetic   Games and   Tricks

A            B                C             D              E
 24,020       36,030           48,040       510,050        612,060

A1           B1               CI            D1             El
 34,212       46,223           58,234       610,245        712,256

A2           B2               C2            D2             E2
 44,404       56,416           68,428       7,104,310      3,124,412

A3           B3               C3            D3             E3
 54,616       66,609          786,112       8,106,215      9,126,318

A4           B4               C4            D4             E4
 64,828      768,112          888,016       9,108,120      10,128,224

A5           B5               C5            D5             E5
750,310      870,215          990,120       10,110,025     11,130,130

A6           B6               C6            D6             E6
852,412      972,318          1,092,224     11,112,130     12,132,036

A7           B7               C7            D7             E7
954,514      1,074,421        1,194,328     12,114,235     13,134,142

A8           B8               C8            D8             E8
1,056,616    1,176,524        1,296,432     13,116,340     14,136,248

A9           B9               C9            D9             E9
1,158,718    1,278,627        1,398,536     14,118,445     15,138,354

  The numbers being very long and 50 in all, your
power will shock all those present. But... you didn't
think you had to learn the 50 long numbers by heart.
Everything is much simpler.
  What is the trickery here?

Another Memory         Thriller
Having written a long series of figures (20 or more),
you proclaim that you can without mistake repeat the
whole series, figure by figure. And really, you put up
a brilliant performance, despite the fact that the
sequence of figures shows no pattern.
  How can you do it?

Mysterious    Cubes
Make several cubes of paper (e.g. four) and write
figures on their faces arranging them as shown in
Fig. 278. With these cubes you can show an interesting
arithmetic trick.
318-319      Arithmetic   Games and   Tricks

Figure 278

               Ask your friends to put the cubes in your absence
             one on top of another in any arrangement to form
             a column. On entering the room you need only cast
             a glance at the column and immediately determine the
             sum of the figures on the closed faces of the four cubes.
             For example, in the case shown in Fig. 278 you would
             call out the sum 23. You can easily see that it is so.

             Trick with Cards
             Make seven cards as shown in Fig. 279. Write the
             numbers on them and cut them exactly as shown. One
             of the cards is left blank, but is cut.
                Now give the six cards with the numbers to your
             friend and ask him to remember one of the numbers
             written on the cards, and then give you back only those
             cards on which there is this number.
                Having received the cards, stack them neatly, put the
             clean card on the top, and add up in your head those
             figures that are seen through the cuts. The result will be
             the number sought.
                You will hardly crack this nut. The trick is based on
             a special selection of numbers in the cards. The idea
             behind it is rather complicated and I am not going to
             dwell on it here.

             How to Find the Sum of Unwritten       Numbers
             You undertake to guess the sum of three numbers of
             which only one is written. The trick is performed as
             follows. Ask your friend to write down any multidigit
             n u m b e r - t h e first summand.
                Suppose he writes 84,706. Then you leave enough
             room for the second and third summands and write
             Arithmetic   Games and   Tricks

Figure 279

             down in advance the sum total of the three numbers:
                                1st summand                  84,706
                                2nd summand
                                3rd summand

                                Sum total                   184,705

               Then your friend writes the second summand (it must
             have the same number of digits as the first), and you
             write the third summand yourself:
                               1st summand                   84,706
                               2nd summand                   30,485
                               3rd summand                   69,514

                               Sum total                    184,705
320-313   Arithmetic   Games and   Tricks

             You can see that the sum was predicted accurately.

          To Foresee a Sum
          In earlier times number superstitions were no less
          widespread than other superstitions. What the result of
          such number fads might be is shown by the example of
          Ilya Teglev, the hero of the story "Rat!...tat!...tat!" by
          Ivan Turgenev. A chance coincidence of numbers led
          him to imagine he was an unrecognized Napoleon.
          After he had committed suicide a sheet of paper was
          found in his pocket with the following calculations:
          Napoleon was born on              Ilya Teglev was born on
          August 15, 1769                   January 7, 1811
          1769                              1811
             8 (August is the 8th month)       1 (January    is   the   1st

          Total 1792                        Total 1819

          Total 19 (!)                      Total 19 (!)

          Napoleon died on                  Ilya Teglev died on
          May 5, 1825                       July 21, 1834
          1825                              1834
             5                                21
                                               7 (July is the 7th month)
              5 (May is the 5th month)
                                            Total 1862
          Total 1835
                  1                                1
                  8                                8
                  3                                6
                  5                                2

          Total 17 (!)                      Total 17(!)

            Such number fortune-telling was widespread at the
          beginning of World War I, when it was hoped the
          outcome could be foreseen using the method. In 1916
          Swiss newspapers initiated their readers into the
          "mysteries" by the following revelation about the fate of
          Emperors of Germany and Austro-Hungary:
   326-313               Arithmetic   Games and   Tricks

                       Wilhelm II        Franz-Joseph
   Year of birth       1859              1830
   Year of accession   1888              1848
   Age                   57                86
   Years reigned         28                68

                Total 3832               3832

   The sums, you see, are equal, each representing the
double of 1916, whence it was concluded that the year
would be fatal for both emperors.
   But here we have not just a chance coincidence, but
human stupidity. Blinded by superstition, the prophets
didn't twig that if you so much as slightly changed the
lines in the calculations, their mysterious character
would go up in smoke.
   Arrange the lines as follows:
   Year of birth
   Year of accession
   Years reigned
   Now what year would you obtain if you add a man's
age to the year of his birth? Of course, the year when
you make your calculation. The same year will result if
to the year of an emperor's accession you add the years
he has reigned. We can easily see now why the adding
up of the four numbers yielded the same result for both
emperors, double 1916. What else could they arrive at?
   You can use this idea for a funny trick. Ask a friend
who doesn't know the trick to write the following four
numbers on a sheet of paper and add them up:
   Year of birth
   Year of ehtering school (factory, etc.)
   Years he's been studying (working, etc.)
   Although you may not know any of the four
numbers it's a simple matter for you to guess their
sum: you only have to double the current year.
   Repeating the trick may well expose the secret. To
muddle up the situation, introduce several additional
numbers you know between the ones you don't. If you
play your cards right, each time the result will be
different and the secret will thus be more difficult to
322-323                         Answers


A Chain of 28 Bones. To simplify the task we'll set aside all the seven doubles: 0-0,
1-1, 2-2, 3-3, 4-4, 5-5, 6-6. The remaining 21 bones have each of the point numbers
repeated six times. For example, the 4-point pattern (on one end) is on the following
six pieces :
                                  4-0, 4-1, 4-2, 4-3, 4-5, 4-6.

   So each number, as we see, occurs an even number of times. Clearly, the pieces of
such a set can be matched to the ends of other pieces until all the set is exhausted.
This done, when the 21 bones are arranged in a continuous line, we insert the
doublets between the butts of 0-0, 1-1, etc. Thus, all the 28 dominoes appear to be
arranged in a line, the rules of the game observed.
Beginning and End of the Chain. We can without difficulty show that the chain of 28
bones must end in the number with which it began. In fact, if this were not the case,
the numbers of points at the ends of the chain would appear an odd number of times
(inside the chain the numbers must occur in pairs). We know, however, that in
a complete set of bones each number occurs eight, i.e. even number of, times.
Consequently, our assumption of unequal point-patterns at the ends of the chain isn't
valid. (An argument like this in mathematics is termed "proof by contradiction".)
   By the way, the property we have just proven suggests the following curious
consequence: a 28-bone chain can always be joined at the ends to yield a ring.
Accordingly, a complete set of dominoes can be arranged not only in a chain with free
ends, but also in a closed ring, all the rules being observed.
   The reader might ask how many ways can such a chain or ring be achieved?
Without launching into the tiresome details of the computation, we will here only say
that this number is e n o r m o u s - 7,959,229,931,520. It represents the product of the
following seven factors: 2 1 3 x 3 8 x 5 x 7 x 4,231.
Trick with Dominoes. The answer follows from what has just been said. We know
that 28 dominoes always make a closed ring, hence if we remove a bone from this
ring, then:
   (1) the remaining 27 dominoes will make a continuous chain with open ends;
   (2) the end numbers in this chain will be those that are on the bone removed.
   Having hidden a bone, we can always predict the point-patterns at the ends of the
chain made up of the remaining bones.
Frame. The points on the sides of the square sought will add up to 44 x 4 = 176, i.e.
8 more than the total in the complete set (168). This, of course, occurs because the
points at the corners of the square are included twice. This yields the sum of points at
the square vertices, i.e. 8. This makes the search for the desired arrangement
somewhat easier, but only slightly. The solution is shown in Fig. 280.
Seven Squares. We'll give two of the many solutions possible. In the first one (at the
top of Fig. 281) we have:
                                1 square   with   sum   3,   2 squares with sum 9,
                                1 square   with   sum   6,   1 square with sum 10,
                                1 square   with   sum   8,   1 square with sum 16.


Figure 280

Figure 281

In the second solution (at the bottom of Fig. 281):
                             2 squares with sum 4,    2 squares with sum 10,
                             1 square with sum 8,     2 squares with sum 12.

Domino Magic Squares. Figure 282 shows an example of the magic square with 18
points in a line.

Figure 282
324-325                       Answers

Domino Progression. By way of example, we'll consider two progressions with
differences equal to 2:
                          (a) 0-0; 0-2; 2-2: 2-4\        4-6.
                          (b)O--l; 1-2; 2-3; 3-4; 4-5; 5-6.
  All told, there are 23 progressions for     6 bones. The initial bones are as follows:
    (a) For unit-difference progressions:
                              0-0, 1-1,       2-1,   2-2, 3-2,
                              0-1, 2-0,       3-0,   3-1, 2-4,
                              1-0, 0-3,       0-4,   1-4, 3-5,
                              0-2, 1-2,       1-3,   2-3, 3-4.
    (b) For progressions with differences of 2:
                             0-0, 0-2, 0-1.
                              The Fifteen      Puzzle
Problem I. The arrangement      can be arrived at        in the   following 44   moves:
                              14, 11, 12, 8, 7, 6, 10,     12, 8, 7,
                               4, 3, 6, 4, 7, 14, 11,      15, 13, 9,
                              12, 8, 4, 10, 8, 4, 14,      11, 15, 13,
                               9, 12, 4, 8, 5, 4, 8,        9, 13, 14,
                              10,   6,   2,   1.

Problem II. The aim is achieved by 39 moves:
                            14, 15, 10, 6, 7, 11, 15, 10, 13, 9,
                             5, 1, 2, 3, 4, 8, 12, 15, 10, 13,
                             9, 5, 1, 2, 3, 4, 8, 12, 15, 14,
                            13, 9, 5, 1, 2, 3, 4, 8, 12.
Problem III. The moves are as follows:
                              12, 8, 4, 3, 2, 6, 10, 9, 13, 15,
                              14, 12, 8, 4, 7, 10, 9, 14, 12, 8,
                               4, 7, 10, 9, 6, 2, 3, 10, 9, 6,
                               5, 1, 2, 3, 6, 5, 3, 2, 1, 13,
                              14, 3, 2, 1, 13, 14, 3, 12, 15, 3.

                              The "11" Game
If you start, you have to take two matches leaving nine. N o matter how many your
partner takes next, you then have to leave only five matches on the table. You should
easily see that you can always do this. And no matter how many of these five your
partner takes, you can leave one match and win.
   If your partner begins, then the outcome of the game depends on whether or not
your partner knows the secret of the fail-safe play.

                                The "15" Game
If you want to win for sure, begin with 5. But in which cell? Let's consider the three
possibilities one by one.
   1. The 5 is written in the middle cell. Which ever cell your opponent chooses, you
can write in the vacant cell in the same row 15-5-x (where x is your opponent's
number). The number 15-5-x, i.e. 10-x, is clearly less than 9.


   2. The 5 is written in a corner cell. Your partner will take either x or y. If he writes
x, you will have to fill in cell y; if he writes y, you respond with x. Either way you win
using the above rule


   3. The 5 is in the middle of the right column. Your partner may occupy one of the
cells: x, y, z, or t.
                                       X   z

                                       y   t

     Your answer to x is t; to y, z; to z, y; to t, x. In all the cases you win.

                                The "32" Game
It's fairly easy to find the way to win in this game, if you take the trouble to play it
backwards from the end. You'll figure out that if your last-but-one draw leaves five
matches on the table, then your win is a sure thing since your partner may not take
more than four matches, hence you can take after him all the remaining matches. But
can you contrive so that you could make your the last but one move leave five
matches on the table? You'll have, by your previous draw, to leave exactly 10
matches, then, whatever his choice, he can't leave you less than six, so that you will
always be able to leave to him five. Further, how can you contrive so that your
partner will have to draw from 10? To achieve this your previous draw must leave 15
matches on the table.
   In this way, by subtracting five each time you'll find that earlier you would have to
leave 20 matches on the table and before that 25 matches, and so at the beginning, 30
matches, i.e. you must begin by drawing 2 matches.
   Thus, for the game to be a success begin by drawing 2 matches, then after your
partner has taken some matches, take as many as are required to leave 25, next go
leave 20, then 10, and finally five. The last match will be yours without fail.
326-327                          nwr
                                A s es

                                The Reverse of the Last Game
Your last-but-one draw must now leave six matches on the table. Then any draw your
partner may make will leave from two to five matches, i.e. your last draw can leave
the last match to him in any event. Thus, your last-but-two draw must leave 11
matches on the table, and on your earlier draw you should leave 16, 21, 26, and 31
matches, respectively.
  You thus begin by taking 1 match and your later draws leave 26, 21, 16, 11, and
6 matches. This will unfailingly leave the last match to your partner.

                                The "27" Game

The way to win here is somewhat more difficult than in the previous game.
   You must start off with the following two considerations:
   1. If before the final draw you have an odd number of matches, you must leave five
matches to your partner and your win is a cinch. In fact, the next draw of your
partner will leave you with four, three, two or one matches. If four are left take three
and win, if three take them and win, if two take one and win, and if one take it and
   2. If before the final draw you have an even number of matches, you must leave six
or seven matches to your partner. In fact, the game will proceed as follows. If your
partner's next draw leaves six matches to you, you must take one and, now with an
odd number of matches, you can safely leave five matches to your partner in which
case he loses all right (see above). If he leaves five matches, not six, you must take four
of them and win; if four take them all and win; if three take two and win; and finally,
if he leaves two you also win. He cannot leave less than two.
   Now you should be able to find the sure way to win without difficulty. If you have
an odd number of matches, you must leave on the table a number of matches that is
a multiple of 6 minus one, i.e. 5, 11, 17, or 23. If you have an even number of
matches, you must leave a multiple of 6 or the multiple plus one, i.e. 6 or 7, 12 or 13,
18 or 19, 24 or 25. Zero is considered an even number, therefore in the beginning you
must take two or three, and then follow the previous procedure.
   If you abide by this rule you will win always. Only you must see to it that your
partner doesn't take the initiative.

                                The Reverse of the Last Game

If you have an even number of matches, you must leave to your partner a multiple of
6 minus one; if you have an odd number, you leave a multiple of 6 or the multiple
plus one. This ensures for your win. At the beginning you have zero matches (i. e. an
even number), therefore your first draw must be to take four matches and leave 23 to
your partner.

                                Think of a Number
No. 1. If the number        thought of is a, then the operations          are as follows
                                (3a + 2) x 3 + a = 10a + 6.

   The result is a two-digit number, the first digit being the number you first thought
of, the second being 6.
   Crossing out the first digit eliminates the number first thought of.
   The rest is self-explanatory.
Nos. 2, 3, 5 and 8 are modifications of what has just been described.
Nos. 4, 6, 7 and 9 use another way of eliminating the number thought of. In No. 9, for
instance, the operations are as follows
                               170 - (a + 20) - 6 + a = 144.
   The rest is self-explanatory.
No. 10 requires a special procedure. To write a three-digit number on the right of
itself is equivalent to multiplying it by 1,001 (e.g. 356 x 1,001 = 356,356). But 1,001 =
 = 7 x 11 x 13. Therefore, if you think of a three-digit number a, then the operations
                                  a x 1,001
                                             = 13.
                                 7 x a x 11

     The rest is clear.

  You thus see that in each of the above cases the guessing is based on eliminating
the number thought of. Now try and devise some new examples of your own.

                               Guessing a Three-Digit     Number

The first digit was first multiplied by 2, then by 5 and by 10, i.e. by 2 x 5 x 10 = 100.
The second digit was multiplied by 10. The third one is added as it is. Besides, we add
5 x 5 x 10, i. e. 250, to the result.
  Thus, if we subtract 250 from the result, we'll have the first digit multiplied by 100
plus the second digit multiplied by 10 plus the third digit. In short, we'll end up with
the number thought of.
  We thus conclude that to guess the number thought of we must subtract 250 from
the result of our calculations.

                               Another Number     Trick

A close examination of the procedure shows that the result must be the four times the
number thought of plus 4. If we thus subtract 4 and divide the rest by 4, we'll arrive at
the number we seek.
                               Guessing the Crossed-Out Digit

Those who know the criterion for divisibility by 9 will know that dividing the sum of
the digits of any number by 9 gives the same remainder as the number itself. Any two
328-329                        Answers

numbers composed of the same digits must therefore give equal remainders when
divided by 9. So if we subtract one of the numbers from the other, the difference will
be exactly divisible by 9 as the subtraction will cancel out the remainders.
  This thus suggests that the digits of the difference your friend obtained add up to
a number divisible by 9. Since the digits 8, 4, 8 that were told to you add up to 20 and
you can infer that the nearest number divisible by 9 is 27 you can find the digit
needed to get 27, and hence the digit crossed-out, which is 7.

                               Guessing the Day and Month of Birth

To work out the date sought we must subtract 365 from the final result. The last two
digits of the difference will then be the number of the month, and the preceding digits
the number of the day. In our example
                               2073 - 365 = 1708.

  From 1708 we determine the date: 17.08. Why?
  Let K be the number of the day, and N the number of the month. We obtain
                           (2K x 10 + 73) x 5 + N = 100K + N + 365.

Clearly, subtracting 365 gives a number with K hundreds and N ones.

                               Guessing Someone's Age

If you go through the procedure several times, you should notice that at all times you
add the age to the same number, namely 1,089. Therefore, if you subtract 1,089 from
the result, you'll obtain the age sought.
   Demonstrating the trick several times you might change the procedure so as not to
expose the secret. For example, by requesting 1,089 be divided by 9 and then add the
age to the result.

                               How Many Sisters? How Many         Brothers?

Subtract 75 from the final result. In our example
                              122 - 75 = 47.

The first digit of the result gives the number of brothers, the second the number of
sisters. In fact, if the number of brothers is a, and the number of sisters is b, then
                               [(a + 3) x (5 x 20)] x 2 + b + 5 = 10a + b + 75

and we arrive at a two-digit number ab.
  The trick can only be a success if the number of sisters is not larger than nine.

                               Trick with a Telephone Directory

The point is that you know the final result beforehand. Whatever the three-digit
number, the outcome is always the same-1,089. You can easily test this. Thus before-
hand you remember the name and number of the subscriber in the ninth line (from
the top or bottom) on page 108.

                               Guessing Domino      Points

Let's trace through the operations to which we subject the first number. We first
multiplied it by 2 and then by 5, i.e. by 10. In addition, we added 7 x 5 = 35.
  Consequently, if we subtract 35 from the result, we'll obtain as many tens as there
are points at one end of the domino. Adding the points at the other end gives the
second digit of the result.
  It's now clear why the figures of the result give the numbers of points at once.

                               Formidable      Memory

The alpha-numerical code of a card indicates the number written on it.
  Above all, you must remember that A stands for 20, B for 30, C for 40, D for 50,
E for 60. Therefore the code means some number. For example Al-21, C3-43, and
  From this number you arrive at the long number written on the card following
a definite rule. Let's discuss it referring to an example.
  Suppose you are told the code E4, i.e. 64. You handle this number as follows:
  First, add up its digits:
                               6 + 4=10.
  Second, double it:
                               64 x 2 = 128.
  Third, subtract the larger digit from the smaller one:
                               6 - 4   = 2.

  Finally, multiply both digits together:
                               6 x 4 = 24.
  Write all the results you obtain in a line
to obtain the number written on the card.
The operations performed could be represented symbolically as
                             + , 2, - , x ,

i.e. adding, doubling, subtracting, multiplying.
330-331                         Answers

  Some more examples:
                  Code D3:                D3 = 53              Code B8: B8 = 38,
                                          5 + 3 = 8,                    3 + 8 = 11,
                                          53 x 2 = 106,                 38 x 2 = 76,
                                          5 - 3 = 2,                    8 - 3 = 5,
                                          5 x 3 = 15.                   8 x 3 = 24.
                                          8,106,215                     1,176,524.
  In order not to strain your memory you can name the numbers as you work them
out or else write them slowly on a blackboard.
  The pattern is rather difficult to discover, therefore the trick generally amazes

                               Another Memory           Thriller
The answer is ridiculously simple: write down the telephone numbers of your

                               Mysterious       Cubes
The answer lies in the arrangement of the numbers on the faces of each cube: the sum
of the numbers on the opposite faces of a cube is seven in all cases (check in Fig. 278).
Therefore, the numbers on the top and bottom faces of all the four cubes stacked in
a column add up to 7 x 4 = 28. If you subtract the number on the top face of the
upper cube from 28, you'll always get the sum of the numbers on all the seven closed
faces of the column.

                               How to Find the Sum of Unwritten           Numbers
If you add 99,999, i.e. 100,000-1, to a five-digit number, then another digit,
1 appears on the left of the number, and the last digit is reduced by 1. The trick is
based on this. So if you mentally add 99,999 to the first number
you can immediately write the future sum of all three numbers, i.e. 184,705. Now you
have only to ensure that the second and third numbers on their own add up to 99,999.
This is achieved by subtracting mentally each digit of the second number from nine
when writing the third number. In our example the second number is 30,485, so you
write 69,514. Since

then the result you've written beforehand will work out without fail.
               With a Stroke of the Pen

               (Drawing figures with one continuous    line)
               The Konigsberg Bridge Problem
               The great mathematician Euler was once interested in
               a curious problem that he described thus:
                 "There is an island called Kneiphof in Konigsberg*.
               The river flowing around it is split into two branches
               which are spanned by seven bridges (Fig. 283).
Figure 2 8 3

                  "Is it possible to visit all of these bridges, crossing
               each once only?
                  "Some people believe that it is possible. Others think
               this is impossible."
                  What do you make of it?

               What is   Topology?
               Euler devoted to the Konigsberg bridge problem
               a whole mathematical investigation that in 1736 he
               presented to the St. Petersburg Academy of Sciences.
               The work was begun with the following words defining
               the branch of mathematics to which similar questions
               might be referred:

                  * Now Kaliningrad in the USSR.
332-333      With a Stroke of the Pen

                "Besides the aspect of geometry that treats of the
             quantities and measuring techniques, an aspect that has
             been developed since ancient times, Leibnitz was the
             first to mention another aspect that he called the
             'geometry of position'. This branch of geometry is only
             interested in the arrangement of the parts of a figure,
             ignoring their sizes*.
                "Recently, I heard of a problem referring to the
             geometry of position, and so I have decided to present
             here by way of example the method I have found of
             solving the problem."
                Euler was referring to the Konigsberg bridge
             problem. We're not now going to discuss the reasoning
             of this eminent mathematician but will only confine
             ourselves to some brief remarks that support his final
             derivation. His conclusion was that it was impossible to
             meet the condition of the problem.

             For simplicity we'll replace the river's branches by the
             scheme in Fig. 284. The size of the island and the
Figure 284   lengths of the bridges are of no consequence and now
             we know this is the characteristic feature of all the
             topological problems.
                Therefore, the localities A, B, C, and D in Fig. 283
             can be replaced by points marked by the same letters
             where the paths meet. Now the problem is seen to
             reduce to tracing the figure in Fig. 284 with one
             continuous path so that no line is drawn twice.
                Let's show that it's impossible to do so. We must
             arrive at each of the node points (A, B, C, and D) by
             one of the paths and then leave by another path. The
             only exception are the initial and final points: since you
             don't corne from anywhere to start and you don't go
             anywhere when you leave. Thus, for our figure to be
             "unicursal" every point, save for two, must meet either
             two, or four (in general any even number) of paths. At
             each of the points A, B, C, and D in the figure odd
             numbers of paths meet. It's thus impossible to trace the
             figure with one continuous path and so it is not
             possible to cross all the Konigsberg bridges as required.

                 * Nowadays this branch of higher geometry is generally termed
             "topology" and it has developed into an extensive field of
             mathematics. The problems in this section of the book belong to
             only a small part of the branch of topology.
             With a Stroke   of the Pen

             Seven Problems
             Try and draw each of the following seven figures with
             one continuous path.

             A Bit of Theory
             Attempts to trace figures 1-6 in Fig. 285 yield different
             results. Some of the figures can be drawn regardless of
             where the path begins. Others can only be drawn if the

Figure 285

             path starts from definite points. Yet others cannot be
             drawn at all by one continuous path. What is the
             reason for this difference? Are there any signs that
             would enable us to predict whether or not a figure is
             unicursal, and if so what must the starting point be?
                The theory provides comprehensive answers to these
             questions, and some elements of the theory will be
             presented below.
                We will now refer to those points at which an even
             number of lines meet as "even" points and to those at
             which an odd number of lines meet as "odd" points.
                It can be shown (we'll omit the proof) that any figure
             only has either zero, or two, or four (in general an even
             number) odd points in it.
                If there are no odd points in the figure, it can always
             be drawn with a single stroke of the pen, wherever you
             start from. Examples are figures 1 and 5 in Fig. 285.
                If there are two odd points in the figure, then it can
             also be drawn in this way, you must only begin from
             one of the odd points (either one). You will find that
334-335        With   a Stroke   of the Pen

               you'll always finish your drawing at the other odd
               point. Examples are figures 2, 3, and 6. In 6, for
               example, you must begin either from point A or from
                  If a figure has more than two odd points, it's
               noncursal. Examples are figures 4 and 7, which both
               contain four odd points.
                  Now you know enough to identify which figures are
               unicursal and the points from which you could start
               your drawing. Professor W. Arens suggests you should
               be guided by another rule, namely "All the lines that
               have already been drawn in a given figure should be
               regarded as absent and when selecting the next line see
               to it that the figure remains complete (doesn't disin-
               tegrate) if the line you've chosen is removed from it."
                  Suppose, for instance, that in figure 5 we've followed
               the path ABCD. If now we draw in line DA, we'll have
               to deal with two figures, ACF and BDE, and they are
               not connected (figure 5 falls apart). Thus, having
               completed figure AFC we won't be able to go over to
               BDE since there'll be no undrawn lines connecting
               them. Therefore, having covered ABCD, you mustn't go
               along DA but should first trace the path DEED and
               then follow the remaining line, DA, over to AFC.

               Seven More         Problems
               Trace figures 8-14 with a continuous line (Fig. 286).

Figure   286
             With a Stroke of the Pen

             The Leningrad     Bridges
             The puzzle is to take a walk around the region of
             Leningrad shown in the figure and come back at the
             starting point whilst crossing each bridge just once.
             Unlike the Konigsberg bridge problem, the task is
             feasible and the reader should now be sufficiently
             armed with knowledge to handle the problem on his

Figure 287
336-337                    Answers


The figures below give the solutions of respective problems in this   section.

Figure 288

Figure 289

22 — 975
             Geometric Recreations

             How Many     Faces?
             How many faces has a hexahedral pencil?
               On the face of it the question is naive, or... intricate.
             Think hard before you look the answer up.

             What Is Shown Here?
             Take a look at Fig. 290. The unusual aspects make
             these objects view outlandish and recognition difficult.
                    ..-- - - m i -   --m     — '— --II " I
Figure 290

             However, try and guess what the figure shows. These
             are all well-known household things.

             Glasses and Knives
             Three glasses are so arranged on the table that their
             mutual separations are larger than the length of a knife

Figure 291
338-339        Geometric   Recreations

               (Fig. 291). Nevertheless, you are asked to contrive
               bridges of these knives such that they connect all three
               glasses. It goes without saying that dislodging the
               glasses is forbidden, as is the use of anything besides
               the three glasses and three knives.
  Figure 292
               How Is It      Achieved?
               You see here (Fig. 292) a wooden cube made up of two
               pieces of wood. The upper half has tongues that fit in
               the grooves of the lower half. But pay attention to their
               shape and arrangement and explain how the joiner
               contrived to connect both parts, each being made of
               a solid block of wood.

               One Plug for Three Holes
               Figure 293 depicts six rows of holes, with three holes in
               each. Using any suitable material, make one plug for
               the three holes of each row.

Figure 293

                  The first row is as easy as pie: clearly the answer is
               the block shown in the figure.
                  As to the other rows the situation is a bit more
               difficult. However, anyone good at engineering drawing
               will make a short work of the task. Essentially, the task
               comes down to manufacturing a component from its
               three views.
 •iVSt,         Geometric      Recreations

                Further "Plug"          Puzzles
Figure    294
                The accompanying figures show three more boards.
                Again, find a plug to close the three holes in each

                Two Cups
                One cup is twice higher than the other, but the other is
                1 1/2 times wider (Fig. 297). Which holds more?

Figure 295      Figure   297

Figure 296

                How Many          Glasses?
                Figure 298 depicts three shelves cn which vessels cf
                three capacities are arranged so that the total capacity
                cf the vessels on each shelf is the same. The smallest
                vessel here is a glass. Find the capacity cf the ether two
                kinds cf vessel.
                Figure   298
340-341        Geometric   Recreations

               Two      Saucepans
               Consider two similar saucepans. Their walls are equally
               thick but one is eight times more capacious than the
Fijvre 299        How many times heavier is it?

               Four Cubes
               Four solid cubes of the same material have different
               heights: 6 cm, 8 cm, 10 cm, and 12 cm (Fig. 299).
               Arrange them on the pans cf a balance for it to be in

               An open barrel contains some water, seemingly half its
               capacity. But you want to know it for certain and you
               don't have a stick or any other measuring device to
               measure the contents of the barrel.
                 Find a way out.

               Which Is Heavier?
               There are two identical cubic boxes (Fig. 300): the one
               on the left contains a large steel ball with a diameter

Figure   300

               equal to the box's height, and the one on the right is
               filled with small steel balls arranged as shown.
                  Which box is heavier?

Figure   301   It's believed that a tripod never rocks, even if its legs
               have different lengths.
                  Is that so?

               How Many        Rectangles?
               How many rectangles can you identify in this figure
               (Fig. 301)? Not squares but rectangles, cf any size.
             346-343             Geometric   Recreations

25 (2>       Chessboard
             How many differently arranged squares could you
             identify on the chessboard?

             A Brick
             A brick weighs 4 kilogrammes.
               What is the weight of a toy brick of the same
             material with all its dimensions four times smaller?

             A Giant and a Dwarf
             Consider a 2-metre giant and a 1-metre dwarf. How
             many times heavier is the giant?

             Along the Equator
Figure 302   If you could walk all the way along the equator, the
             top of your head would have travelled a longer way
             than each point on your feet.
                What would this difference be?

             Through a Magnifying      Glass
             An angle of 1 1/2° is viewed through a 4 x power
             magnifying glass (Fig. 302).
               What will its apparent magnitude be?

             Similar Figures
             This problem is for those who know about the concept
             of geometric similarity. Referring to Fig. 303, answer

Figure 303

                          a                           b
             the following questions:
                1. Are the internal and external triangles similar
             (Fig. 303a)?
                2. In the frame of the picture (Fig. 3036), are the
             internal and external rectangles similar?
342-343              Geometric   Recreations

                     The Height of a Tower
                     Suppose there is a tourist attraction in your town,
                     a tower whose height you don't know, however. But
                     you have got a photograph of the tower on a picture
                       How could this picture help you to determine the

                     A Strip
                     A bit of mental arithmetic: if a square metre is divided
                     into 1-mm squares, and all of them are arranged side
                     by side on a straight line, how long will be the strip
Figure   304
                     A Column
                     Now imagine a column produced by stacking all the
                     1-mm cubes contained in 1 cubic metre. How high
                     would this column be?
                     Which is heavier: a glassful-of granulated sugar or
                     pressed sugar?

                     The Path of a Fly
                     Consider a cylindrical glass jar 20 centimetres high and
                     10 centimetres in diameter. On the inner wall,
                     3 centimetres from the top, there is a drop of honey,
Figure   305         and on the outer wall, the diametrically opposite, there
                     is a fly (Fig. 304).
                        Trace the shortest path for the fly to get to the
                        Don't hope that the fly could find the shortest way
                     on its own, thereby simplifying the problem. This
                     would require a knowledge of gebmetry on its part,
                     that would be "superflyish".

                     The Path of a Beatle
                     At the roadside lies      a granite block 30 centimetres long,
                     20 centimetres high       and 20 centimetres thick (Fig. 305).
                     A beatle is sitting       at point A and wants to find the
                     shortest way to B.
                       Trace the path           and   find out   how   long it is.
348-343           Geometric   Recreations

A Bumble-Bee's   Travels
A bumble-bee sets out on a long journey. From its nest
it flies due south, crosses a river and after an hour's
travel alights on a hill covered with clover. Fluttering
from a flower to flower it spends half-hour here.
   Now it goes to the orchard where the bumble-bee
yesterday saw gooseberry-bushes in blossom. The
orchard lies due west of the hill and it makes
a "bumble-bee" line there, where it arrives 3/4 of an
hour later. The bushes being in full blossom, the insect
takes 1 1/2 hours to visit all of them.
   At last, the bumble-bee starts on its return journey
and takes the shortest route possible.
   How long has the bumble-bee been away from its

The Foundation of Carthage
According to a tradition concerning the foundation of
Carthage the Tyrian princess Dido, who lost her
husband at the hands of her brother, fled to the north
coast of Africa with many of the inhabitants of Tyre.
She bought from the Numidian king as much land "as
an oxen hide occupies". Having concluded the bargain
Dido had the hide cut into thin belts and thanks to this
trick she got a site big enough for a fortress to be
erected. So the citadel of Carthage was built, and later
developed into a city.
   How calculate the area that, according to the legend,
the fortress could occupy, given that the oxen hide had
a surface area of 4 square metres and the belts into
which Dido had it cut were 1 millimetre wide.
344-345                         Answers

                       CJ)      How Many     Faces?
The problem reveals an incorrect usage of words. A hexagonal pencil doesn't have six
faces, as may well be believed. If it isn't sharpened, it has eight faces, all in all: six
lateral faces and two small "end" faces. If it really had six faces, it would have quite
another shape, namely a block with a rectangular cross section.
   The habit of only counting side faces in prisms, and ignoring the end faces is
widespread. Many people say: trihedral prisms, tetrahedral prisms, etc., whereas these
prisms should be referred to as triangular prisms, quadrangular prisms, etc., according
to their cross section. What is more, a trihedral prism (i. e. having three faces) cannot
   The pencil mentioned in the problem should be referred to as "hexagonal", not
                                What Is Shown Here?
The objects are a razor, a pair of scissors, a fork, a pocket watch, and a spoon. When
we look at some object we, generally speaking, see it projected onto a plane normal to
the line of sight. Here you were not shown the views that you see habitually and this
is enough to render an object almost unrecognizable.

                                Glasses and Knives
This is quite possible to achieve by arranging the knives as shown in Fig. 306.
Figure 306

                                                      Figure 307

                               How Is It    Achieved?
The way out is very simple, as can be seen from Fig. 307.
2 3 - 975

     3                       One Plug for Three Holes
The suitable plugs are shown in Fig. 308.

Figure 308

                             Further "Plug"    Puzzles
In this case, the plugs are more complicated (Figs. 309, 310, 311).

Figure 309                                  Figure   310

Figure 311
346-347                         Answers

                                Two Cups
The cup that is 1 1/2 times wider would (with the same height) have (1 1/2)2, i.e. 2 1/4
times more volume. Since it is only half the height of the other cup, in the final
analysis it still holds more than the taller cup.

                                How Many      Glasses?
A comparison of the first and third shelves shows that they differ only in that the
third shelf contains one more middle-sized vessel whilst the three small vessels are
missing. The total capacity of the vessels on each shelf being the same, it's obvious
that the capacity of one middle-sized vessel equals that of the three small ones. The
middle-sized vessel thus equals three glasses. It only remains now to determine the
capacity of a large vessel. By replacing all the middle-sized vessels on the first shelf by
the appropriate number of glasses we get one large vessel and 12 glasses on the top
   Comparison with the second half yields that one large vessel holds 6 glasses.

                                Two Saucepans
The two saucepans are geometrically similar. Given that the larger saucepan holds
8 times more, all of its dimensions are twice larger: it's twice higher and wider. Its
surface area must then be 2 x 2 times larger, because the surfaces of similar bodies
relate to each other as the squares of their linear dimensions. The thickness of walls
being the same, the weight of a saucepan depends on its surface area. The answer is
therefore that the larger pan is four times heavier.

                                Four Cubes
We must place the three smaller cubes on one pan, and the largest one on the other.
It's easily verified that the balance will be in equilibrium. Let's show that the total
volume of the three smaller cubes equals that of the largest one. This follows from the
                                63 + 83 + 103 = 123,
i. e.
                                216 + 5 1 2 + 1,000= 1,728.

The simplest way is to tilt the barrel so that the water reaches the edge (Fig. 312). If
some of the bottom shows above the surface, however little, the barrel is less than
half-full. If, on the contrary, the bottom is well below the surface of the water, the
barrel is more than half-full. Finally, if the upper edge of the bottom is exactly on the
water level, the barrel is exactly half-full.

Figure 312

                               Which Is Heavier?
Let's imagine the right cube as consisting of small cubes, each containing a ball. It's
easily seen that the large ball occupies the same proportion of the large cube's volume
as each small ball occupies of the smaller cube's volume. We can readily work out the
number of these small balls and cubes: 6 x 6 x 6 = 216. The total volume of the 216
balls accounts for the same share of the 216 cubes as the big ball relative to the big
cube. It follows that both boxes contain the same amount of metal, and hence their
weight is the same.

A tripod can always touch the floor with each of its three legs, because through any
three points in space one can draw a plane, and only one at that. This explains why
a tripod doesn't rock. This problem, you see, is purely geometrical and not physical.
  That is why tripods are so convenient as supports for field instruments and cameras.
A fourth leg wouldn't make the support any more stable.

                              How Many      Rectangles?

The chessboard contains more than 64 squares. Apart from the small black and white
squares there are the larger squares consisting of 4, 9, 16, 25, 36, 49, and 64 unit
squares. These must be taken into account as well.

                               Unit Squares           Number     on
                                   i                     64
                                   4                     49
                                   9                     36
                                  16                     25
                                  25                     16
                                  36                      9
                                  49                      4
                                  64                      1

                                              Total     204

  Thus, the chessboard contains 204 differently arranged squares of different sizes.

                               A Brick
If you thought the toy brick weighs 1 kilogramme, i.e. only a quarter lighter,      you
would be wrong. It's not only a quarter the length, but also a quarter the width    and
a quarter the height of a standard brick, therefore its volume and weight is 4 x    4x
 x 4 = 64 times less.
   Consequently, the correct answer is: the toy brick weighs 4,000 — 64 =           62.5

                               A Giant and a Dwarf
Now you are well equipped to solve this problem correctly. Since human bodies are
approximately similar, the giant would be eight times heavier, not twice as heavy.
  The tallest giant ever recorded was a man from Alsace in Germany. He was 275
centimetres high, a metre higher than an average man. The smallest dwarf was under
40 centimetres, i.e. he was seven times smaller than the Alsatian giant. Therefore, if
the giant were to stand on one pan of a balance, 7 x 7 x 7 = 343 or a whole crowd of
dwarfs would have to stand on the other to balance.

                               Along the Equator
We take the man to be 175 centimetres high and denote the Earth's radius by R. We
thus have
                                2 x 3.14 x (R + 175) - 2 x 3.14 xR = 2x 3.14 x 175 =
                               = 1,100 cm,
i.e. about 11 metres. Interestingly enough, the result is independent of the globe's

                               Through a Magnifying           Glass
If you believe that the magnifying glass will make the angle look as if it were 1 1/2° x
 x 4 = 6°, you put your foot in it. Viewing through a magnifying glass doesn't make

the angle any larger. True, the arc subtending the angle increases, but the radius of the
arc increases as much, with the result that the central angle remains the same (Fig.
Figure   313

                                Similar Figures
Not infrequently, both questions are answered in the affirmative. In actual fact, only
the triangles are similar. For triangles to be similar it's sufficient for the angles to be
equal and since the sides of the inner triangle are parallel to those of the external one,
the angles are equal. With other polygons it's not sufficient only to have equal angles
(or parallel sides which is mathematically the same). It is also necessary that their sides
be proportional. For the internal and external rectangles of the frame this is only the
case for squares (more generally, for rhombs). In any other cases, however, the sides of
the external rectangle are not proportional to the sides of the internal rectangle, and
hence the figures are not similar. This stands out especially for rectangular frames with

Figure   314

wide planks as shown in Fig. 314. In the left frame the ratio of the external sides is
2:1, and of the inner sides 4 : 1 . In the right frame the ratio of the external sides is
4 : 3 , and of the internal sides 2 : 1 .

                                The Height of a Tower
To work out the height of the tower we should at first measure as accurately as
possible the height of the tower and the length of its base in the picture. Suppose the
height in the picture is 95 centimetres and the base is 19 centimetres. Now measure
the base of the real tower, which is, say, 14 metres.
  Then you argue as follows. The picture and the real tower are similar, the height-
to-base ratio in reality and in the picture are the same. The first ratio is 95:19 = 5,
then you conclude that the height of the real tower is five times larger than its base:
14 x 5 = 70 metres, i.e. the tower is 70 metres high. It's worth noting that the method
350-351                         Answers

only works with pictures that don't distort proportions which is often the case with
inexperienced cameramen.

                               A Strip
There are 1,000,000 square millimetres in 1 metre. Each thousand 1-mm squares
arrange along a line span 1 metre, and a thousand thousand 1-mm squares give 1,000
metres, i.e. 1 kilometre. Thus the strip will be 1 kilometre long.

                               A Column
The answer is striking: the column will be... 1,000 kilometres high.
   Let's test it mentally. There are 1000 x 1000 x 1000 cubic millimetres in 1 cubic
metre. Each thousand 1-mm cubes stacked one upon another gives a 1-metre column.
Multiplied by 1,000 this gives 1,000 metres = 1 kilometre. Multiplying by the last
1,000, we obtain 1,000 kilometres.

Use your imagination. Suppose for simplicity that the lumps of pressed sugar are 100
times larger across than the particles of granulated sugar. Now imagine that all the
granules in the granulated sugar were enlarged 100 times together with the glass
containing them. The capacity of the glass would be increased 100 x 100 x 100, i.e.
one million times, as would the weight of the sugar. Let's take a normal glassful of
this enlarged granulated sugar, i.e. one millionth part of the contents of the giant
glass. Ciearly, it will weigh as much as a normal glassful of conventional granulated
sugar. But then, what is the enlarged granulated sugar? It's nothing but pressed sugar.
Accordingly, a glassful of pressed sugar has the same weight as that of granulated
  If we had made the magnification 60-fold instead of 100-fold or any other mag-
nification, the situation wouldn't have changed in the least. The key thing here is that
the pieces of pressed sugar are regarded here as being geometrically similar to the
particles of granulated sugar and are at that arranged in a similar manner. The
assumption is not strict, but it's fairly close to reality if the lumps are irregular.

                               The Path of a Fly
Let's make the sides of the cylinder jar into a flat surface. We'll obtain a rectangle
(Fig. 315a) 20 centimetres high with the base equal to the circumference of the jar, i.e.
10 x 3 1/7 = 311/2 centimetres (approximately). On this rectangle we now can mark
the positions of the fly (A) and honey drop (B).
  Now to find the point at which the fly must cross the edge we'll proceed as follows.
We'll draw a line from B (Fig. 3156) at a right angle to the upper side of the rectangle
and continue it an equal distance beyond the edge. We obtain point C which we
connect with a line to A. Point D will be where the fly must cross the edge to the
other side, the path ADB being the shortest.
                                 tfm Answers

Figure 315

  Having found the shortest way on the rectangle, we'll again make it into a cylinder
and find out how our fly must walk to get to the honey drop in the shortest time
possible (Fig. 3156).

                                The Path of a Beatle
We'll mentally turn the upper face of the stone so that it lies in the same plane as the
front face (Fig. 316). The shortest route then is the line connecting A and B. What is

Figure 316


its length? We have the right triangle ABC, where AC = 40 cm, CB = 30 cm.
According to the Pythagorean theorem, AB must be 50 cm, because 30 2 + 40 2 = 502.
So the shortest path AB = 50 cm.

                                A Bumble-Bee's      Travels
The problem would be a "piece of cake", if we knew the time taken by the bumble-bee
to cover the distance from the orchard to its nest. Geometry will help us work this
   Let's draw the path of the insect. We know that it flew at first "due south" for 60
minutes. Then it flew for 45 minutes "due west", i. e. at right angles to the first leg, and
finally it flew back to its nest by the shortest path possible, i.e. along a straight line.
We thus obtain the right triangle ABC with two known legs, AB and BC, the hypote-
nuse AC remaining to be determined.
352-353                        Answers

Figure   317


                                 Or char c
                                                             45 min

   Geometry teaches that if one leg of a right-angled triangle is three units long, the
other leg four units long, then the hypotenuse is exactly five units long.
   For example, if legs are 3 metres and 4 metres, then the hypotenuse is 5 metres; if
9 and 12 kilometres, then the hypotenuse is 15 kilometres, and so forth. In our case
one leg is 3 x 15 minutes of flight long and the other 4 x 15 minutes long, hence the
hypotenuse AC = 5 x 15 minutes of flight long. We have thus found that the
bumble-bee took 75 minutes, i.e. 11/4 hour, to cover the distance from the orchard to
its nest.
   Now it's child's play to figure out how long our bumble-bee had been away from its
                                 Flights: 1 + 3 / 4 + 1 1 / 4 = 3 hours.
                                 Stops: 1/2 + 1 1 / 2 = 2 hours.
                                 Total: 3 + 2 = 5 hours.

                                 The Foundation of Carthage
Since the surface area of the hide was 4 square metres, or 4 million square millimetres,
and the belt thickness was 1 millimetre, the total length of the belt (clearly, Dido had
it cut in a spiral) was 4 million millimetres, or 4,000 metres, i.e. 4 kilometres long.
A belt this long can encircle a square area of 1 square kilometre, or a round area of
1.3 square kilometres.
Without a Tape-Measure

Measuring by Paces*
Since a tape-measure is not always at hand, it pays to
be able to do without one where approximate estimates
are sufficient.
   Longish distances, for instance during hikes, can be
conveniently measured by paces. But this does of
course require that we know how long our paces are
and could count them. Admittedly, paces are not
always the same: we can walk with short steps or long
steps, when we need to. But still when we walk at
a measured pace our steps are about similar, and if we
know their average length, we can without much error
measure distances in paces.
   To find the average length of your pace you should
measure the total length of many paces and find the
length of one. Here, of course, we cannot do without
a tape-measure.
   Lay out the tape on a smooth piece of ground and
measure a 20-metre stretch. Draw in the line and
remove the tape. Now walk along the line in your
normal way and count the number of paces you have.
It's possible that the stretch of ground you measured
does not contain an integer number of paces. Then, if
the remainder is shorter than a step, it can be simply
discarded; if it's longer than a step, the remainder is
taken to be a whole step. Dividing the total length of
20 metres by the number of paces gives the average
length of one pace. This number should be remembered
so that, if necessary, it might be used for measuring.
   In order not to lose count of paces you can,
especially over long distance, use the following trick.
Count up to 10 and then tick off a finger of your left
hand. After the five fingers of the left hand have been
ticked off, i.e. 50 paces covered, start ticking off the
fingers on the right hand. We can in this way count up
to 250, and then start from the very beginning
remembering how many times have we ticked off all the
fingers of the right hand. For example, if having
covered a certain distance you've ticked off all the
fingers of the right hand twice and you end up with
three fingers more ticked off on the right hand, and

  * We will call two steps 1 pace.
354-355        Without   a   Tape-Measure

               four on the left, you've made
               2 x 250 + 3 x 50 + 4 x 10 = 690 paces.
                  You should also add the paces you made after the
               last finger of your left hand has been ticked off.
                  By the way, there is an old rule which says that the
               length of an average step of an adult equals the
               distance from floor to his eyes.
                  Another old practical rule refers to walking speed:
               a man covers as many kilometres in half an hour as he
               makes paces in 3 seconds. We can easily show that the
               rule is only true for one rather large length of pace. Let
               the .pace length be x metres, and the number of paces
               made in 3 seconds be n. Then in 3 seconds the walker
               goes nx metres, and in half an hour (1,800 seconds) he
               travels 600 nx metres, or 0.6 nx kilometres. For this
               distance to be equal to the number of paces made in
               3 seconds the following equality should hold
               0.6 nx = n, or 0.6 nx = 1.
               Hence x = 1.66 metres.
                 If the first old rule relating pace length to man's
               height, then the second rule is only valid for people of
               average height, about 175 centimetres.

               Living Scales
               To measure objects appropriately without a tampe-
               measure, you can proceed as follows. Measure by

Figure   318
             360-355            Without   a   Tape-Measure

             a stick or a rope the length from the end of your
             outstretched    arm     to   your    opposite    shoulder
             (Fig. 318)-in adult it's about a metre. Another way of
             getting an approximate metre is to mark off six times
             the distance from your thumb to forefinger, separated
             as widely as possible (Fig. 319a).
                The latter piece of advice introduces us to the art of
             measuring "with bare hands". Initially, you should only
             measure parts of your hand and remember the results.
               Which parts then should be measured? Above all, its
             width, as shown in Fig. 319b. In adults it's about 10
             centimetres, but it varies from person to person and
Figure 319

             you should know its exact value. Then it pays to know
             the span between the ends of your middle and index
             fingers separated as wide as possible (Fig. 319c). It is
             also advisable to know the length of your index finger
             from the base of your thumb, as shown in Fig. 319d
             and, finally, the width of your outspread palm from
             thumb to little finger, as shown in Fig. 319e.
                This "live scale" will enable you to estimate the
             dimensions of small objects.

             Measuring with Coins
             It pays to know the size of your national coins, because
             they might be of help in measuring objects. By way of
             example, the Soviet coins have convenient dimensions:
356-355        Without   a   Tape-Measure

               1 kopeck piece is exactly 11/2 centimetres across,
               5 kopeck piece is 21/2 centimetres across, and so on.
               Remember the diameters of your coins!
Figure   320
                  Simple Tricks and Diversions

                  Guessing Domino Points
                  The trick is based on a dodge that can't be guessed.
                    You could surprise your friends by saying that you'll
                  guess the drawn domino's points from an adjoining
                  room. For better effect suggest they blindfold you.
                  Really, one of your friends draws the piece and asks
                  you to guess its points and you, from the adjacent
                  room, give them the right numbers straight away and
                  without so much as a glance at it or your friends.
                    What is the idea behind the trick?

                  Disappearing Line
                  Copy out the figure in Fig. 321 very accurately. Cut the
                  ring out, apply it to the ring in the figure and turn it
                  counterclockwise so that the severed part of each line
Figure 321        registers with the remains of a neighbouring one. You'll
                  witness something enigmatic: instead of the 13 lines
                  that were there before the figure will only show 12. One
                  line will have disappeared. Where to?
                     The reverse operation brings the line back. Where

                  A Mysterious   Knot
                  We'll now turn to trick with things.
                     Here is a curious trick that could surprise your
                     Take a piece of string about 30 centimetres long
                  (Fig. 322) and make a loose knot on it as shown on the
Figure 322

                  left of the accompanying figure. Add a second loop (see
                  the knot in the middle). You're sure to expect that
                  tightening the string now will give you a good double
                  knot. But to be on the safe side we'll make the knot
358-359        Simple   Tricks   and   Diversions

               smarter by threading one of the loose ends through
               both loops as shown on the right.
                  All the preparations over, we can proceed to the
               main part of the trick. Take hold of one end of the
               string and offer the other to your friend. Pull! You'll
               discover something neither you nor your friend
               expected: instead of an involved knot you'll have
               a smooth piece of string! The knot will have gone.
                  The trick is a success if only you make the third loop
               exactly as shown. So examine the knots in the figure

               Bind your friends (A and B) as shown in Fig. 323.
                 Is it possible to set the friends free without cutting
               the strings?

Figure   323


               A Pair of Boots
               Take a sheet of strong paper and cut out a frame,
               a pair of boots and an oval ring as shown in Fig. 324.
               The hole in the oval ring is the size of the width of the
Figure   324
             Simple   Tricks   and   Diversions

             frame, but narrower than the legs of the boots. There-
             fore, if you are asked to hang the boots on the frame as
             shown in the figure, you'll obviously think that it's
Figure 325      But it is possible. How?

             Corks on a Ring
             There is a ring of strong paper, on which two corks
             hang suspended from a short piece of string with a wire
             ring slung on the string as shown in Fig. 325.
                Remove the corks from the paper ring.

Figure 326
             Two      Buttons
             Figure 326 shows a sheet of paper with two long cuts
             and one small oval hole that is a bit smaller than the
             separation between the long cuts. Thread a piece of
             string through the hole and the cuts and tie a button to
             each end of the string so that the buttons won't pass
             through the hole.

             "Magic      Purse"
             Cut two rectangles out of a sheet of cardboard, the
             rectangles being the size of a notebook, say
             7 centimetres long and 5 centimetres wide. Get three
             pieces of ribbon (paper strips will do as well), two of
             them being a centimetre longer than the rectangles'
             width, and the third a centimetre longer than twice the
             width of the rectangles. Glue the ribbons to the
Figure 327   rectangles as shown in Fig. 327. In so doing, bend the
             ends of the shorter ribbons under the right rectangle
             and glue them to it, and glue the other ends to the
             back side of the left rectangle. Glue the end of the
             longer ribbon to the outside of the right rectangle,
             thread the ribbon under it, then round the outside of
             the left rectangle and glue its end under this rectangle.
                You now have your "magic" purse. Using it, you can
             show your friends a fascinating trick that can be
             dubbed "live paper" or something like that. Take
             a piece of paper signed by your friend so that you
             could not replace it. Stick the paper under the two
             ribbons. Now close the purse, reopen it. Presto! The
             paper itself emerged from under both papers but (what
360-359    Simple   Tricks    and   Diversions

           is beyond belief!) it got under the centre ribbon on the
           opposite side of the purse.

           Guessing          Matches
           In my childhood I was much amazed by a trick shown
           to me by my elder brother. Going about my business in
           my room once I heard in the adjacent room some
           laughter that wetted my curiosity. I peered in and saw
           my brother and his student friend laughing.
              "Come in, boy! We'll show you an interesting trick."
              That was exactly what I wanted. My brother was
           a great wag.
              "Look here," said Alex arranging matches on a table
           in a random manner, "I put ten matches down at
           random. Now Til go into the kitchen and you think of
           a match here. When you're ready, call me. I'll just take
           a look at the matches and tell you which one it was."
              "And he'll say that it's not the right one," the guest
           interrupted, "No, some control is needed here, we can't
           do without it!"
              "Okay, we'll do it this way: when he's thought of
           a match he'll show it to you. You'll be a witness."
              "That's different. Let's start."
              My brother left. I made sure that he was gone and
           couldn't peer into the keyhole. Then I thought of
           a match, showed it to the student without touching and
           called out: "Ready!"
              I didn't believe that Alex would guess the match
           since I hadn't so much as touched it and all the
           matches remained in their places. How could he
           possibly guess?
              But he did! He just came up to the table and without
           a moment's hesitation pointed to the match. I even
           tried hard not to look at it in order not to betray
           myself. My brother even didn't glance at me and still
           guessed... Well I never!
              "Want another go?"
              "Of course!"
              We did it again and he guessed it again! A dozen
           times we did the trick and each time my brother
           indicated the match I had thought of without mistake.
           I was on the verge of bursting into tears as I was dying
           to know the secret. Finally, my tormentors took pity on
           me and revealed the trick.
              What was it?
24   975
               366-359   Simple   Tricks   and   Diversions

               Eleven Matches        on One
               Arrange a dozen matches as shown in Fig. 328 and try
               and raise them all by lifting the sticking out end of the

Figure   328

               lower match. If you are adroit enough, the trick will
               work, if not, practice a bit.

               Is It   Easy?
               What do you make of what is shown in Fig. 329: is it
               easy to lift a match with two other matches?
Figure   329

                 It seems easy as pie, doesn't it? But try to do it
               yourself and you'll find that it requires patience and
               practice, the slightest jolt will turn the match over.

               On a Narrow         Path
               Draw a narrow path of 15 squares on a sheet of paper
               (Fig. 330).
                  For the game you'll need a die and two counters or
               draughts (two coins or buttons will also do).
                  The rules of the game are simple. Each of the two
               partners places his counter at the either end of the
               path. Then they take turns to throw the die (the one
               with the largest number of points begins). Each partner
               shifts his counter forward by so many squares as there
               are points shown by the die, but he is not entitled to
362-363        Simple   Tricks    and   Diversions

Figure   330

               skip the square occupied by his opponent's counter. If
               the die shows more points than there are free squares
               left, he must retreat by the excess number of squares.
                  The counters thus alternatively appear in the middle
               of the path or at its extremes. The game ends when one
               of the partners is forced to leave the path. The winner
               is the one who stays.

               Star-Like         Patterns
               Some people maybe don't know that just with a pair of
               scissors, without any drawing instruments it's possible
               to manufacture an infinite variety of beautiful paper
                  Take a sheet of white paper and fold it several times
               as shown in Fig. 331, A, B, C, D, and E. Having

Figure   331

               reached the stage E, cut the folded paper along some
               ornate lines, e. g. like those shown in the figure.
                 Now unfold and smooth out the paper to obtain
             368-359   Simple   Tricks and   Diversions

             a beautiful design that will look yet better when glued
             on some dark paper (Fig 332).
Figure 332

             Five-Pointed       Star
             Can you cut out a paper five-pointed star? It is not
             simple and takes some practice, otherwise your star will
             have unequal points. There are two methods of cutting
             good regular stars.
                In the first method, a circle is drawn on a sheet of
             paper using a pair of compasses or just a saucer. The
             circle is cut out and folded in two, the semicircle
             obtained is then folded four times as shown in
             Fig. 333/4.
Figure 333

                This is the most difficult part of the problem as it
             requires a good eye, because the semicircle must be
             folded so as to give five similar segments.
                Once the circle has been folded correctly, it is
             trimmed at the thick end along one of the dash lines in
             Fig. 333B. When you unfold the paper, you get
             a regular five-pointed star with either shallow or deep
             notches (Fig. 333, C and D) depending on how you
             trimmed the semicircle.
                The second method is perhaps simpler as we start
             with a square, not a circle. To begin with, a square
             sheet of paper (Fig. 334/1) is folded in two. Then three
364-359        Simple   Tricks   and   Diversions

               more folds are made as shown in Fig. 334, B, C, and D.
               The dot-and-dash line in Fig. 334D indicates the trim
               line. The resultant star is depicted in Fig. 334£.

Figure   334
             Drawing Puzzles

             What's Written   Here?
             Something is written in the circle (Fig. 335). Looking at
             it in the conventional way, you will, of course, perceive

Figure 335

             nothing sensible. However, if you view the circle in the
             proper way, you'll be able to read the words. Which
             words ?

             Its Simple. Or Is It?
             Look carefully at the design in Fig. 336 and try to
             remember it so that you could reproduce it from
             memory. Have you remembered it?... Then begin
             drawing. At first, mark out the four end points of the
Figure 336

             two lines. The first curve will probably come out
             adequately. Okay! Now draw the second curve. But the
             line is stubbornly unsuccess. This seemingly easy job
             does not now appear to be so easy.

             On Which Foot
             Look at Fig. 337 and say which leg the footballer is
             standing on, the right or left.
                He seems to be standing on his right leg, but you can
             say that he is standing on his left leg with the same
366-367      Drawing   Puzzles

Figure 337

             measure of certainty. No matter how long you view the
             drawing you'll never answer the question. The artist has
             done his job so skillfully that it's impossible to establish
             which leg is doing the kick and which is supporting the
               Perhaps you are asking, "But which is which, really?"
             I don't know. The artist doesn't know, either. It will
             remain an unsolvable mystery for ever.

             How Many        Fish?
             You see a strange drawing here (Fig. 338). It might
             seem that the angler has caught nothing so far. But

Figure 338

             look very attentively at the figure: three big fish are
             already here. Where are they?
               Drawing   Puzzles

               Where Is the Tamer?
               Where is the tamer of this tiger (Fig. 339)?
                His portrait does appear in this figure. Find it.

Figure   339

               Look at the picture (Fig. 340)-a sunset-and say if it is
                  The picture contains one incongruity that you should

Figure   340
368-369      Drawing   Puzzles

             Figure 341 represents a tropical moonset. Is the picture
               Perhaps you can see something incongruous about it.

Figure 341
                        tfm    Answers

                   27   €
                        6      Guessing Domino Points
Here you use a secret language known only to you and one of your friends with
whom you've preliminarily worked it out. You've agreed, say, that the following words
have the meaning:
                             "I", and "my" - 1
                             "you", and " y o u r " - 2
                             "he", and " h i s " - 3
                             "we", and " o u r " - 4
                             "they", and "their"-5
                             "it", and " i t s " - 6
   These conventions may be illustrated by some examples. Let the piece in question
be 4-6. In that case, your companion calls out:
   "We've thought of a piece, guess it."
   In the secret language this will be: "we"-4, " i t " - 6 , hence 4-6.
   If the piece is 1-5, then your companion utters:
   "I think this time they are difficult to guess."
   Those uninitiated will never guess that these words contain the secret message:
' T - l , "they"-5.
   A further example: 4-2. What "message" should your companion send? Something
like this:
   "Now we've thought of such a piece that you'll never guess."
   But how about the blanks? It might be denoted by some other word, say, friend. If
the domino is 0-4, your fellow-conspirator calls out:
   "Guess, friend, what we've thought now."

                               Disappearing Line
The jist of the trick is better illustrated in a simplified form. Figure 342 shows a piece
of cardboard with 13 lines. The sheet is cut along the diagonal. If you shift one part
relative to the other as shown in the figure, then instead of 13 lines you'll get only 12,

Figure   342

one will have disappeared. In this case we can easily see where it has gone since each
of the 12 new lines has got somewhat longer than before, namely a 1/12th longer.
370-371                        Answers

Clearly, when shifted one of the lines has been divided into 12 parts each of which
went to lengthen the other lines. Reverse shifting brings the vanished line back into
being by shortening the other lines.
  The lines in Fig. 321 are arranged in a circle and possess the same property:
shifting the circle through an appropriate angle kills one of the lines (it is "smeared"
over the other 12).

Yes, it is.
  String A is taken at point C and threaded through loop B in the direction indicated

Figure   343


by the arrow. When a sufficient length of the string has already been tucked in, hand
B is put into the loop formed and, when string A is pulled, the friends separate.

                               A Pair of Boots
The accompanying figure explains the answer. The frame is folded and the ring is put
on the folded ends. Then the unfolded figure of "double boots" is threaded in-between
the folded ends, refolded and pushed to the bend in the frame. Finally, the ring is
slided onto the end. It now only remains to unfold the frame.

Figure   344

     6                        Corks on a Ring
Now that you know how to solve the previous problem, this one will be smooth

Figure   345

  Fold the paper ring as shown, remove the wire ring by sliding it away to the free
end, and remove the corks.

                              Two       Buttons
The accompanying figure shows the solution. Fold the paper so that the upper and

Figure   346

lower ends of the narrow strip between the cuts will coincide. Then thread the strip
through the oval hole and remove the buttons through the loop.

                              "Magic      Purse"
The point is that you open the purse from the opposite side.

                               Guessing      Matches
I was simply made a fool of. The student who pretended to control the guessing was
my brother's conspirator and gave signals to him.
   But how? That was the trick of it. It turned out that the matches were arranged not
at random: the brother had so arranged them (Fig. 347) that the pattern would
resemble the outlines of a human face. So the upper match marked the hair, the next
below, the forehead, further down were the eyes, nose, mouth, chin, neck, and on
either side the ears. When Alex walked into the room, he first of all cast a glance at
372-371                        Answers

Figure   347

the "controller" who touched an appropriate feature of his face with his hand, thus
indicating which match had been thought of.

                               What's Written   Here?
Bring the ring up to your eyes as shown in Fig. 348. You'll clearly read the words
MIR PUBLISHERS, then turn the circle and you'll now see the word PERELMAN.

Figure   348

  The letters are extremely elongated and narrow, therefore it's impossible to make
them out in the conventional way. In the suggested method the letters become much
shorter, their width being the same. This imparts a normal aspect to tht letters, thus
simplifying reading them.

                               How Many     Fish?
Fll help you discern the catch. One fish is one the angler's back, head down. Another
is between the float and the end of the fishing rod. The third is under his feet.

                               Where Is the Tamer?
The tiger's eye doubles as the tamer's eye, the tamer is looking in the opposite

The incongruity is that the convex part of the crescent faces in the opposite direction
from the sun, not towards it. The moon is illuminated by the sun, hence it by no
means can be facing the sun with it dark side...
  The French astronomer Flammarion wrote: "Most of painters are ignorant of this,
because a year never passes without a large number of inverted crescents appearing at
the Paris Salon".

'                              Moonset
Strange as it may seem, the crescent in Fig. 341 is depicted correctly. It's a tropical
landscape where the position of the crescent differs from that in the higher latitudes,
where the hump of the new moon faces to the right and that of the old moon faces to
the left. But in tropical lands the crescent hangs horizontally in the sky.
   This is explained as follows. In the higher latitudes the sun and moon (indeed all
the luminaries in general) execute their diurnal motion in inclined circles. Therefore
during the evening the sun casts slanting rays at the moon, illuminating it from the
right or left, so that the crescent faces to the right or left. But on the equator the
celestial bodies move in normal trajectories with the result that the sun illuminating
the moon sets below the horizon directly beneath the moon and not to the left or
right of it. The moon is thus illuminated from below and that is why the crescent has
the form of a gondola, as shown in the figure.
T h e End

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